GSEB Class 11 Maths Solutions Chapter 4 Principle of Mathematical Induction Exercise 4.1

Question 1. Prove the following by using the principle of mathematical induction for all n ∈ N? \( 1 + 3 + 3^2 + \dots + 3^{n-1} = \frac{3^n-1}{2} \).
Answer: Let P(n) represent the provided statement, which is:
\( P(n) : 1 + 3 + 3^2 + \dots + 3^{n-1} = \frac{3^n-1}{2} \)
By setting \( n = 1 \), we find that \( P(1) = \frac{3-1}{2} = 1 \).
Therefore, P(n) holds for \( n = 1 \).
Now, let's assume that P(k) is true for some positive integer k.
So, we can write: \( P(k) : 1 + 3 + 3^2 + \dots + 3^{k-1} = \frac{3^k-1}{2} \)
We will now demonstrate that P(k + 1) is also true whenever P(k) holds.
To do this, we add \( 3^k \) to both sides of the equation:
\( 1 + 3 + 3^2 + \dots + 3^{k-1} + 3^k = \frac{3^k-1}{2} + 3^k \)

\( \implies \frac{3^k-1 + 2 \cdot 3^k}{2} \)

\( \implies \frac{3^k (1 + 2) - 1}{2} \)

\( \implies \frac{3 \cdot 3^k - 1}{2} \)

\( \implies \frac{3^{k+1} - 1}{2} \)
This outcome matches the formula for P(k + 1).
Thus, P(k + 1) is also true whenever P(k) holds.
Hence, by the principle of mathematical induction, P(n) is true for all natural numbers \( n \in N \).
In simple words: We showed the formula works for the first number, then assumed it works for any number 'k'. By adding the next term and simplifying, we proved it also works for 'k+1', which means the formula is true for all natural numbers.

Exam Tip: Remember to clearly state the base case (n=1), the inductive hypothesis (assuming P(k) is true), and the inductive step (proving P(k+1) is true) in your solution to get full marks.

Question 2. Prove the following by using the principle of mathematical induction for all n ∈ N? \( 1^3 + 2^3 + 3^3 + \dots + n^3 = \left(\frac{n(n+1)}{2}\right)^2 \).
Answer: Let P(n) represent the statement:
\( P(n) : 1^3 + 2^3 + 3^3 + \dots + n^3 = \left(\frac{n(n+1)}{2}\right)^2 \)
For \( n = 1 \):
L.H.S. \( = 1^3 = 1 \)
R.H.S. \( = \left(\frac{1(1+1)}{2}\right)^2 = \left(\frac{1 \times 2}{2}\right)^2 = 1^2 = 1 \)
Since L.H.S. = R.H.S., P(1) is true.
Assume that P(k) holds for some positive integer k.
So, we have: \( P(k) : 1^3 + 2^3 + 3^3 + \dots + k^3 = \left(\frac{k(k+1)}{2}\right)^2 \) (2)
We now aim to prove that P(k + 1) is true, assuming P(k) is true.
Add \( (k+1)^3 \) to both sides of equation (2):
\( 1^3 + 2^3 + 3^3 + \dots + k^3 + (k+1)^3 = \left(\frac{k(k+1)}{2}\right)^2 + (k+1)^3 \)

\( \implies \frac{k^2(k+1)^2}{4} + (k+1)^3 \)

\( \implies (k+1)^2 \left(\frac{k^2}{4} + (k+1)\right) \)

\( \implies (k+1)^2 \left(\frac{k^2 + 4(k+1)}{4}\right) \)

\( \implies (k+1)^2 \left(\frac{k^2 + 4k + 4}{4}\right) \)

\( \implies (k+1)^2 \frac{(k+2)^2}{4} \)

\( \implies \frac{(k+1)^2(k+2)^2}{4} \)

\( \implies \left(\frac{(k+1)(k+2)}{2}\right)^2 \)

\( \implies \left(\frac{(k+1)((k+1)+1)}{2}\right)^2 \)
This is the expression for P(k + 1).
Therefore, P(k + 1) holds for \( n = k + 1 \).
Hence, by the principle of mathematical induction, P(n) is true for all natural numbers \( n \in N \).
In simple words: We showed that if the sum of cubes formula works for 'k', it also works for 'k+1'. Since it works for 1, it must work for all natural numbers.

Exam Tip: When performing algebraic manipulations in the inductive step, look for common factors like \( (k+1)^2 \) to simplify expressions and show the form for P(k+1).

Question 3. Prove the following by using the principle of mathematical induction for all n ∈ N? \( 1 + \frac{1}{1+2} + \frac{1}{1+2+3} + \dots + \frac{1}{1+2+3+\dots+n} = \frac{2n}{n+1} \).
Answer: Let P(n) represent the statement:
\( P(n) : 1 + \frac{1}{1+2} + \frac{1}{1+2+3} + \dots + \frac{1}{1+2+3+\dots+n} = \frac{2n}{n+1} \)
For \( n = 1 \):
L.H.S. \( = 1 \)
R.H.S. \( = \frac{2 \cdot 1}{1+1} = \frac{2}{2} = 1 \)
Since L.H.S. = R.H.S., P(1) is true.
Assume that P(k) holds for some positive integer k.
So, we have: \( P(k) : 1 + \frac{1}{1+2} + \frac{1}{1+2+3} + \dots + \frac{1}{1+2+3+\dots+k} = \frac{2k}{k+1} \)
We now aim to demonstrate that P(k + 1) is true, assuming P(k) is true.
Add the \( (k+1)^{th} \) term to both sides. The \( (k+1)^{th} \) term is \( \frac{1}{1+2+3+\dots+(k+1)} = \frac{1}{\frac{(k+1)(k+2)}{2}} = \frac{2}{(k+1)(k+2)} \).
So, L.H.S. becomes:
\( 1 + \frac{1}{1+2} + \frac{1}{1+2+3} + \dots + \frac{1}{1+2+3+\dots+k} + \frac{1}{1+2+3+\dots+(k+1)} \)
Using P(k), this is: \( \frac{2k}{k+1} + \frac{2}{(k+1)(k+2)} \)

\( \implies \frac{2k(k+2) + 2}{(k+1)(k+2)} \)

\( \implies \frac{2k^2 + 4k + 2}{(k+1)(k+2)} \)

\( \implies \frac{2(k^2 + 2k + 1)}{(k+1)(k+2)} \)

\( \implies \frac{2(k+1)^2}{(k+1)(k+2)} \)

\( \implies \frac{2(k+1)}{k+2} \)

\( \implies \frac{2(k+1)}{(k+1)+1} \)
This is the expression for P(k + 1).
Therefore, P(k + 1) holds whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all natural numbers \( n \in N \).
In simple words: We showed that the formula holds for 'n=1'. Then, we assumed it works for 'k' and used that to prove it also works for 'k+1', confirming it's true for all natural numbers.

Exam Tip: In problems involving sums, always remember the formula for the sum of the first n natural numbers, \( \sum_{i=1}^n i = \frac{n(n+1)}{2} \), as it's frequently used to simplify terms.

Question 4. Prove the following by using the principle of mathematical induction for all n ∈ N? \( 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + 3 \cdot 4 \cdot 5 + \dots + n(n+1)(n+2) = \frac{n(n+1)(n+2)(n+3)}{4} \).
Answer: Let P(n) represent the statement:
\( P(n) : 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + 3 \cdot 4 \cdot 5 + \dots + n(n+1)(n+2) = \frac{n(n+1)(n+2)(n+3)}{4} \)
For \( n = 1 \):
L.H.S. \( = 1 \cdot 2 \cdot 3 = 6 \)
R.H.S. \( = \frac{1(1+1)(1+2)(1+3)}{4} = \frac{1 \cdot 2 \cdot 3 \cdot 4}{4} = 6 \)
Since L.H.S. = R.H.S., P(1) is true.
Assume that P(k) holds for some positive integer k.
So, we have: \( P(k) : 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + 3 \cdot 4 \cdot 5 + \dots + k(k+1)(k+2) = \frac{k(k+1)(k+2)(k+3)}{4} \) (2)
We now aim to demonstrate that P(k + 1) is true, assuming P(k) is true.
Add the \( (k+1)^{th} \) term, which is \( (k+1)(k+2)(k+3) \), to both sides of equation (2):
L.H.S. \( = 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + 3 \cdot 4 \cdot 5 + \dots + k(k+1)(k+2) + (k+1)(k+2)(k+3) \)
Using P(k), this is: \( \frac{k(k+1)(k+2)(k+3)}{4} + (k+1)(k+2)(k+3) \)

\( \implies (k+1)(k+2)(k+3) \left(\frac{k}{4} + 1\right) \)

\( \implies (k+1)(k+2)(k+3) \left(\frac{k+4}{4}\right) \)

\( \implies \frac{(k+1)(k+2)(k+3)(k+4)}{4} \)

\( \implies \frac{(k+1)((k+1)+1)((k+1)+2)((k+1)+3)}{4} \)
This is the expression for P(k + 1).
Therefore, P(k + 1) holds whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all natural numbers \( n \in N \).
In simple words: We confirmed the formula works for 'n=1'. Then, we showed that if it works for 'k', it must also work for 'k+1', proving it's true for all natural numbers.

Exam Tip: When dealing with products or sums that involve sequential terms, factoring out common factors in the inductive step can significantly simplify the algebra and lead to the desired P(k+1) form.

Question 5. Prove the following by using the principle of mathematical induction for all n ∈ N? \( 1 \cdot 3 + 2 \cdot 3^2 + 3 \cdot 3^3 + \dots + n \cdot 3^n = \frac{(2n-1) \cdot 3^{n+1}+3}{4} \).
Answer: Let P(n) represent the statement:
\( P(n) : 1 \cdot 3 + 2 \cdot 3^2 + 3 \cdot 3^3 + \dots + n \cdot 3^n = \frac{(2n-1) \cdot 3^{n+1}+3}{4} \)
For \( n = 1 \):
L.H.S. \( = 1 \cdot 3 = 3 \)
R.H.S. \( = \frac{(2 \cdot 1-1) \cdot 3^{1+1}+3}{4} = \frac{1 \cdot 3^2+3}{4} = \frac{9+3}{4} = \frac{12}{4} = 3 \)
Since L.H.S. = R.H.S., P(1) is true.
Assume that P(k) holds for some positive integer k.
So, we have: \( P(k) : 1 \cdot 3 + 2 \cdot 3^2 + 3 \cdot 3^3 + \dots + k \cdot 3^k = \frac{(2k-1) \cdot 3^{k+1}+3}{4} \) (1)
We now aim to demonstrate that P(k + 1) is true, assuming P(k) is true.
Add the \( (k+1)^{th} \) term, which is \( (k+1) \cdot 3^{k+1} \), to both sides of equation (1):
L.H.S. \( = 1 \cdot 3 + 2 \cdot 3^2 + \dots + k \cdot 3^k + (k+1) \cdot 3^{k+1} \)
Using P(k), this is: \( \frac{(2k-1) \cdot 3^{k+1}+3}{4} + (k+1) \cdot 3^{k+1} \)

\( \implies \frac{(2k-1) \cdot 3^{k+1} + 3 + 4(k+1) \cdot 3^{k+1}}{4} \)

\( \implies \frac{3^{k+1} (2k-1 + 4(k+1)) + 3}{4} \)

\( \implies \frac{3^{k+1} (2k-1 + 4k+4) + 3}{4} \)

\( \implies \frac{3^{k+1} (6k+3) + 3}{4} \)

\( \implies \frac{3^{k+1} \cdot 3(2k+1) + 3}{4} \)

\( \implies \frac{3^{k+2} (2k+1) + 3}{4} \)

\( \implies \frac{(2(k+1)-1) \cdot 3^{(k+1)+1} + 3}{4} \)
This is the expression for P(k + 1).
Therefore, P(k + 1) holds whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all natural numbers \( n \in N \).
In simple words: We confirmed the formula works for the first case (n=1). Then, we showed that if it holds for 'k', it must also hold for 'k+1', proving its validity for all natural numbers.

Exam Tip: Be careful with exponent rules and factoring common terms, especially when dealing with \( 3^{k+1} \) and \( 3^{k+2} \), as a small error can lead to an incorrect P(k+1) form.

Question 6. Prove the following by using the principle of mathematical induction for all n ∈ N? \( 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + n(n+1) = \frac{n(n+1)(n+2)}{3} \).
Answer: Let P(n) represent the statement:
\( P(n) : 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + n(n+1) = \frac{n(n+1)(n+2)}{3} \)
For \( n = 1 \):
L.H.S. \( = 1 \cdot 2 = 2 \)
R.H.S. \( = \frac{1(1+1)(1+2)}{3} = \frac{1 \cdot 2 \cdot 3}{3} = 2 \)
Since L.H.S. = R.H.S., P(1) is true.
Assume that P(k) holds for some positive integer k.
So, we have: \( 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + k(k+1) = \frac{k(k+1)(k+2)}{3} \)
The last term in L.H.S. is \( k(k+1) \).
Replacing k with k + 1, the next term becomes \( (k+1)((k+1)+1) = (k+1)(k+2) \).
We will now prove that P(k + 1) is true by adding \( (k+1)(k+2) \) to both sides:
L.H.S. \( = 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + k(k+1) + (k+1)(k+2) \)
Using P(k), this becomes: \( \frac{k(k+1)(k+2)}{3} + (k+1)(k+2) \)

\( \implies (k+1)(k+2) \left(\frac{k}{3} + 1\right) \)

\( \implies (k+1)(k+2) \left(\frac{k+3}{3}\right) \)

\( \implies \frac{(k+1)(k+2)(k+3)}{3} \)

\( \implies \frac{(k+1)((k+1)+1)((k+1)+2)}{3} \)
This is the expression for P(k + 1).
Thus, P(k + 1) holds whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all natural numbers \( n \in N \).
In simple words: We established the formula for 'n=1'. Then, we showed that if it works for 'k', it also works for 'k+1', which means it's true for all natural numbers.

Exam Tip: When the terms involve products of consecutive integers, look for ways to factor out the common product of terms in the inductive step to simplify the expression efficiently.

Question 7. Prove the following by using the principle of mathematical induction for all n ∈ N? \( 1 \cdot 3 + 3 \cdot 5 + 5 \cdot 7 + \dots + (2n-1)(2n+1) = \frac{n(4n^2+6n-1)}{3} \).
Answer: Let P(n) represent the given statement:
\( P(n) : 1 \cdot 3 + 3 \cdot 5 + 5 \cdot 7 + \dots + (2n-1)(2n+1) = \frac{n(4n^2+6n-1)}{3} \)
For \( n = 1 \):
L.H.S. \( = 1 \cdot 3 = 3 \)
R.H.S. \( = \frac{1(4 \cdot 1^2+6 \cdot 1-1)}{3} = \frac{4+6-1}{3} = \frac{9}{3} = 3 \)
Since L.H.S. = R.H.S., P(1) is true.
Assume that P(k) holds for some positive integer k.
So, we have: \( P(k) : 1 \cdot 3 + 3 \cdot 5 + 5 \cdot 7 + \dots + (2k-1)(2k+1) = \frac{k(4k^2+6k-1)}{3} \)
The last term in L.H.S. is \( (2k-1)(2k+1) \).
Replacing k with k + 1, the next term will be \( (2(k+1)-1)(2(k+1)+1) = (2k+1)(2k+3) \).
We will now prove that P(k + 1) is true by adding \( (2k+1)(2k+3) \) to both sides:
L.H.S. \( = 1 \cdot 3 + 3 \cdot 5 + 5 \cdot 7 + \dots + (2k-1)(2k+1) + (2k+1)(2k+3) \)
Using P(k), this becomes: \( \frac{k(4k^2+6k-1)}{3} + (2k+1)(2k+3) \)

\( \implies \frac{k(4k^2+6k-1) + 3(2k+1)(2k+3)}{3} \)

\( \implies \frac{4k^3+6k^2-k + 3(4k^2+8k+3)}{3} \)

\( \implies \frac{4k^3+6k^2-k + 12k^2+24k+9}{3} \)

\( \implies \frac{4k^3+18k^2+23k+9}{3} \)
We need to show this is equal to \( \frac{(k+1)(4(k+1)^2+6(k+1)-1)}{3} \).
Let's expand the target R.H.S. for P(k+1):
\( \frac{(k+1)(4(k^2+2k+1)+6k+6-1)}{3} \)
\( \implies \frac{(k+1)(4k^2+8k+4+6k+5)}{3} \)
\( \implies \frac{(k+1)(4k^2+14k+9)}{3} \)
\( \implies \frac{4k^3+14k^2+9k+4k^2+14k+9}{3} \)
\( \implies \frac{4k^3+18k^2+23k+9}{3} \)
Since the simplified L.H.S. matches the expanded R.H.S. for P(k+1),
P(k + 1) holds whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all natural numbers \( n \in N \).
In simple words: We verified the formula for 'n=1'. Then, by assuming it works for 'k' and adding the next term, we showed it also works for 'k+1', proving its truth for all natural numbers.

Exam Tip: For complex algebraic expressions in the inductive step, it can be helpful to expand the target P(k+1) form separately and then try to match your derived expression to it. This provides a clear goal for your manipulation.

Question 8. Prove the following by using the principle of mathematical induction for all n ∈ N? \( 1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + \dots + n \cdot 2^n = (n-1)2^{n+1}+2 \).
Answer: Let P(n) represent the statement:
\( P(n) : 1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + \dots + n \cdot 2^n = (n-1)2^{n+1}+2 \)
For \( n = 1 \):
L.H.S. \( = 1 \cdot 2 = 2 \)
R.H.S. \( = (1-1)2^{1+1}+2 = 0 \cdot 2^2 + 2 = 0 + 2 = 2 \)
Since L.H.S. = R.H.S., P(1) is true.
Assume that P(k) holds for some positive integer k.
So, we have: \( P(k) : 1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + \dots + k \cdot 2^k = (k-1)2^{k+1}+2 \)
The last term in L.H.S. is \( k \cdot 2^k \).
Replacing k with k + 1, the next term will be \( (k+1) \cdot 2^{k+1} \).
We will now prove that P(k + 1) is true by adding \( (k+1) \cdot 2^{k+1} \) to both sides:
L.H.S. \( = 1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + \dots + k \cdot 2^k + (k+1) \cdot 2^{k+1} \)
Using P(k), this becomes: \( (k-1)2^{k+1}+2 + (k+1)2^{k+1} \)

\( \implies 2^{k+1} (k-1 + k+1) + 2 \)

\( \implies 2^{k+1} (2k) + 2 \)

\( \implies k \cdot 2 \cdot 2^{k+1} + 2 \)

\( \implies k \cdot 2^{k+2} + 2 \)
This is the expression for P(k + 1).
Thus, P(k + 1) holds whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all natural numbers \( n \in N \).
In simple words: We confirmed the formula for 'n=1'. Then, by assuming it works for 'k' and adding the next term, we showed it also works for 'k+1', proving its truth for all natural numbers.

Exam Tip: When simplifying, look for common exponential terms like \( 2^{k+1} \) to factor out. This helps in combining terms and reaching the desired P(k+1) form more easily.

Question 9. Prove the following by using the principle of mathematical induction for all n ∈ N? \( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^n} = 1 - \frac{1}{2^n} \).
Answer: Let P(n) represent the statement:
\( P(n) : \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^n} = 1 - \frac{1}{2^n} \)
For \( n = 1 \):
L.H.S. \( = \frac{1}{2} \)
R.H.S. \( = 1 - \frac{1}{2^1} = 1 - \frac{1}{2} = \frac{1}{2} \)
Since L.H.S. = R.H.S., P(1) is true.
Assume that P(k) holds for some positive integer k.
So, we have: \( P(k) : \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^k} = 1 - \frac{1}{2^k} \)
We now aim to demonstrate that P(k + 1) is true, assuming P(k) is true.
Add the \( (k+1)^{th} \) term, which is \( \frac{1}{2^{k+1}} \), to both sides:
L.H.S. \( = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^k} + \frac{1}{2^{k+1}} \)
Using P(k), this becomes: \( 1 - \frac{1}{2^k} + \frac{1}{2^{k+1}} \)

\( \implies 1 - \frac{2}{2 \cdot 2^k} + \frac{1}{2^{k+1}} \)

\( \implies 1 - \frac{2}{2^{k+1}} + \frac{1}{2^{k+1}} \)

\( \implies 1 - \frac{2-1}{2^{k+1}} \)

\( \implies 1 - \frac{1}{2^{k+1}} \)
This is the expression for P(k + 1).
Therefore, P(k + 1) holds whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all natural numbers \( n \in N \).
In simple words: We showed the formula works for the first number, then assumed it works for any number 'k'. By adding the next term, we proved it also works for 'k+1', which means the formula is true for all natural numbers.

Exam Tip: When combining fractional terms with different denominators involving powers, find a common denominator (usually the higher power) to simplify correctly.

Question 10. Prove the following by using the principle of mathematical induction for all n ∈ N? \( \frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \dots + \frac{1}{(3n-1)(3n+2)} = \frac{n}{6n+4} \).
Answer: Let P(n) represent the statement:
\( P(n) : \frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \dots + \frac{1}{(3n-1)(3n+2)} = \frac{n}{6n+4} \)
For \( n = 1 \):
L.H.S. \( = \frac{1}{2 \cdot 5} = \frac{1}{10} \)
R.H.S. \( = \frac{1}{6 \cdot 1+4} = \frac{1}{10} \)
Since L.H.S. = R.H.S., P(1) is true.
Assume that P(k) holds for some positive integer k.
So, we have: \( P(k) : \frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \dots + \frac{1}{(3k-1)(3k+2)} = \frac{k}{6k+4} \)
The \( k^{th} \) term is \( \frac{1}{(3k-1)(3k+2)} \).
The \( (k+1)^{th} \) term is \( \frac{1}{(3(k+1)-1)(3(k+1)+2)} = \frac{1}{(3k+3-1)(3k+3+2)} = \frac{1}{(3k+2)(3k+5)} \).
We now aim to demonstrate that P(k + 1) is true by adding the \( (k+1)^{th} \) term to both sides:
L.H.S. \( = \frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \dots + \frac{1}{(3k-1)(3k+2)} + \frac{1}{(3k+2)(3k+5)} \)
Using P(k), this becomes: \( \frac{k}{6k+4} + \frac{1}{(3k+2)(3k+5)} \)

\( \implies \frac{k}{2(3k+2)} + \frac{1}{(3k+2)(3k+5)} \)

\( \implies \frac{1}{3k+2} \left(\frac{k}{2} + \frac{1}{3k+5}\right) \)

\( \implies \frac{1}{3k+2} \left(\frac{k(3k+5) + 2}{2(3k+5)}\right) \)

\( \implies \frac{1}{3k+2} \left(\frac{3k^2+5k+2}{2(3k+5)}\right) \)
Factor the numerator \( 3k^2+5k+2 = (3k+2)(k+1) \).

\( \implies \frac{1}{3k+2} \left(\frac{(3k+2)(k+1)}{2(3k+5)}\right) \)

\( \implies \frac{k+1}{2(3k+5)} \)

\( \implies \frac{k+1}{6k+10} \)

\( \implies \frac{k+1}{6(k+1)+4} \)
This is the expression for P(k + 1).
Therefore, P(k + 1) holds whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all natural numbers \( n \in N \).
In simple words: We proved the formula for the first term. Then, we showed that if it holds for 'k', it also works for 'k+1', which means it's true for all natural numbers.

Exam Tip: When working with sums of fractions, always aim to factor out common terms from the denominator and find a common denominator to simplify the expression. Factoring quadratic expressions in the numerator is also a key step.

Question 11. Prove the following by using the principle of mathematical induction for all n ∈ N? \( \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{3 \cdot 4 \cdot 5} + \dots + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)} \).
Answer: Let P(n) represent the statement:
\( P(n) : \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{3 \cdot 4 \cdot 5} + \dots + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)} \)
For \( n = 1 \):
L.H.S. \( = \frac{1}{1 \cdot 2 \cdot 3} = \frac{1}{6} \)
R.H.S. \( = \frac{1(1+3)}{4(1+1)(1+2)} = \frac{1 \cdot 4}{4 \cdot 2 \cdot 3} = \frac{4}{24} = \frac{1}{6} \)
Since L.H.S. = R.H.S., P(1) is true.
Assume that P(k) holds for some positive integer k.
So, we have: \( P(k) : \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{3 \cdot 4 \cdot 5} + \dots + \frac{1}{k(k+1)(k+2)} = \frac{k(k+3)}{4(k+1)(k+2)} \)
The \( k^{th} \) term is \( \frac{1}{k(k+1)(k+2)} \).
The \( (k+1)^{th} \) term is \( \frac{1}{(k+1)((k+1)+1)((k+1)+2)} = \frac{1}{(k+1)(k+2)(k+3)} \).
We now aim to demonstrate that P(k + 1) is true by adding the \( (k+1)^{th} \) term to both sides:
L.H.S. \( = \frac{1}{1 \cdot 2 \cdot 3} + \dots + \frac{1}{k(k+1)(k+2)} + \frac{1}{(k+1)(k+2)(k+3)} \)
Using P(k), this becomes: \( \frac{k(k+3)}{4(k+1)(k+2)} + \frac{1}{(k+1)(k+2)(k+3)} \)

\( \implies \frac{1}{(k+1)(k+2)} \left(\frac{k(k+3)}{4} + \frac{1}{k+3}\right) \)

\( \implies \frac{1}{(k+1)(k+2)} \left(\frac{k(k+3)^2 + 4}{4(k+3)}\right) \)

\( \implies \frac{1}{(k+1)(k+2)} \left(\frac{k(k^2+6k+9) + 4}{4(k+3)}\right) \)

\( \implies \frac{1}{(k+1)(k+2)} \left(\frac{k^3+6k^2+9k+4}{4(k+3)}\right) \)
Let's analyze the numerator \( k^3+6k^2+9k+4 \). We know that \( (k+1) \) must be a factor for the final form. If \( k = -1 \), then \( -1+6-9+4 = 0 \), so \( (k+1) \) is a factor.
Using polynomial division or synthetic division, we get \( k^3+6k^2+9k+4 = (k+1)(k^2+5k+4) = (k+1)(k+1)(k+4) = (k+1)^2(k+4) \).

\( \implies \frac{1}{(k+1)(k+2)} \left(\frac{(k+1)^2(k+4)}{4(k+3)}\right) \)

\( \implies \frac{(k+1)(k+4)}{4(k+2)(k+3)} \)

\( \implies \frac{(k+1)((k+1)+3)}{4((k+1)+1)((k+1)+2)} \)
This is the expression for P(k + 1).
Therefore, P(k + 1) holds whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all natural numbers \( n \in N \).
In simple words: We confirmed the formula for the first term. Then, we showed that if it holds for 'k', it also works for 'k+1', which means it's true for all natural numbers.

Exam Tip: For problems with complex polynomial numerators, remember to factorize them, often looking for \( (k+1) \) as a factor. This simplifies the expression and helps in achieving the P(k+1) form.

Question 12. Prove the following by using the principle of mathematical induction for all n ∈ N? For \( r \neq 1 \), \( a + ar + ar^2 + \dots + ar^{n-1} = \frac{a(1-r^n)}{1-r} \).
Answer: Let P(n) represent the statement:
\( P(n) : a + ar + ar^2 + \dots + ar^{n-1} = \frac{a(1-r^n)}{1-r} \)
For \( n = 1 \):
L.H.S. \( = a \)
R.H.S. \( = \frac{a(1-r^1)}{1-r} = \frac{a(1-r)}{1-r} = a \)
Since L.H.S. = R.H.S., P(1) is true.
Assume that P(k) holds for some positive integer k.
So, we have: \( P(k) : a + ar + ar^2 + \dots + ar^{k-1} = \frac{a(1-r^k)}{1-r} \) (2)
We now aim to demonstrate that P(k + 1) is true, assuming P(k) is true.
Add the \( (k+1)^{th} \) term, which is \( ar^k \), to both sides of equation (2):
L.H.S. \( = a + ar + ar^2 + \dots + ar^{k-1} + ar^k \)
Using P(k), this becomes: \( \frac{a(1-r^k)}{1-r} + ar^k \)

\( \implies \frac{a(1-r^k) + ar^k(1-r)}{1-r} \)

\( \implies \frac{a - ar^k + ar^k - ar^{k+1}}{1-r} \)

\( \implies \frac{a - ar^{k+1}}{1-r} \)

\( \implies \frac{a(1 - r^{k+1})}{1-r} \)
This is the expression for P(k + 1).
Therefore, P(k + 1) holds whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all natural numbers \( n \in N \).
In simple words: We confirmed the sum of geometric series formula for the first term. Then, we assumed it works for 'k' and added the next term, showing it works for 'k+1', thus proving it for all natural numbers.

Exam Tip: This is the formula for the sum of a geometric progression. Make sure you understand the base case and how adding the \( (k+1)^{th} \) term leads directly to the next form of the sum.

Question 13. Let the given statement be denoted by P(n), i.e., \( P(n) : \left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)\dots\left(1+\frac{2n+1}{n^2}\right) = (n+1)^2 \).
Answer: Let P(n) represent the given statement:
\( P(n) : \left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)\dots\left(1+\frac{2n+1}{n^2}\right) = (n+1)^2 \)
For \( n = 1 \):
L.H.S. \( = \left(1+\frac{2 \cdot 1+1}{1^2}\right) = \left(1+\frac{3}{1}\right) = 4 \)
R.H.S. \( = (1+1)^2 = 2^2 = 4 \)
Since L.H.S. = R.H.S., P(1) is true.
Assume that P(k) holds for some positive integer k.
So, we have: \( P(k) : \left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)\dots\left(1+\frac{2k+1}{k^2}\right) = (k+1)^2 \)
We now aim to demonstrate that P(k + 1) is true, assuming P(k) is true.
Multiply both sides by the \( (k+1)^{th} \) term, which is \( \left(1+\frac{2(k+1)+1}{(k+1)^2}\right) = \left(1+\frac{2k+3}{(k+1)^2}\right) \):
L.H.S. \( = \left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)\dots\left(1+\frac{2k+1}{k^2}\right)\left(1+\frac{2k+3}{(k+1)^2}\right) \)
Using P(k), this becomes: \( (k+1)^2 \left(1+\frac{2k+3}{(k+1)^2}\right) \)

\( \implies (k+1)^2 \left(\frac{(k+1)^2 + 2k+3}{(k+1)^2}\right) \)

\( \implies (k+1)^2 + 2k+3 \)

\( \implies k^2+2k+1 + 2k+3 \)

\( \implies k^2+4k+4 \)

\( \implies (k+2)^2 \)

\( \implies ((k+1)+1)^2 \)
This is the expression for P(k + 1).
Therefore, P(k + 1) holds whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all natural numbers \( n \in N \).
In simple words: We showed the statement is true for n=1. Then, by assuming it holds for 'k' and multiplying by the next term, we proved it also holds for 'k+1', confirming its truth for all natural numbers.

Exam Tip: When dealing with products, the inductive step involves multiplying both sides by the next term. Carefully expand and simplify the quadratic expressions to match the P(k+1) form.

Question 14. Let the given statement be P(n), i.e., \( P(n) : \left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\dots\left(1+\frac{1}{n}\right) = n+1 \).
Answer: Let P(n) represent the given statement:
\( P(n) : \left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\dots\left(1+\frac{1}{n}\right) = n+1 \)
For \( n = 1 \):
L.H.S. \( = \left(1+\frac{1}{1}\right) = 2 \)
R.H.S. \( = 1+1 = 2 \)
Since L.H.S. = R.H.S., P(1) is true.
Assume that P(k) holds for some positive integer k.
So, we have: \( P(k) : \left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\dots\left(1+\frac{1}{k}\right) = k+1 \)
We now aim to demonstrate that P(k + 1) is true, assuming P(k) is true.
Multiply both sides by the \( (k+1)^{th} \) term, which is \( \left(1+\frac{1}{k+1}\right) \):
L.H.S. \( = \left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\dots\left(1+\frac{1}{k}\right)\left(1+\frac{1}{k+1}\right) \)
Using P(k), this becomes: \( (k+1) \left(1+\frac{1}{k+1}\right) \)

\( \implies (k+1) \left(\frac{k+1+1}{k+1}\right) \)

\( \implies (k+1) \left(\frac{k+2}{k+1}\right) \)

\( \implies k+2 \)
This is the expression for P(k + 1).
Therefore, P(k + 1) holds whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all natural numbers \( n \in N \).
In simple words: We confirmed the statement for the base case (n=1). Then, by assuming it works for 'k' and multiplying by the next term, we proved it also holds for 'k+1', which means it's true for all natural numbers.

Exam Tip: Simplify terms like \( \left(1+\frac{1}{n}\right) = \frac{n+1}{n} \) before substitution into the inductive step, as this often makes the cancellation process much clearer.

Question 15. Let the given statement be P(n), i.e., \( P(n) : 1^2 + 3^2 + 5^2 + \dots + (2n-1)^2 = \frac{n(2n-1)(2n+1)}{3} \).
Answer: Let P(n) represent the given statement:
\( P(n) : 1^2 + 3^2 + 5^2 + \dots + (2n-1)^2 = \frac{n(2n-1)(2n+1)}{3} \)
For \( n = 1 \):
L.H.S. \( = 1^2 = 1 \)
R.H.S. \( = \frac{1(2 \cdot 1-1)(2 \cdot 1+1)}{3} = \frac{1 \cdot 1 \cdot 3}{3} = 1 \)
Since L.H.S. = R.H.S., P(1) is true.
Assume that P(k) holds for some positive integer k.
So, we have: \( P(k) : 1^2 + 3^2 + 5^2 + \dots + (2k-1)^2 = \frac{k(2k-1)(2k+1)}{3} \)
We now aim to demonstrate that P(k + 1) is true, assuming P(k) is true.
Add the \( (k+1)^{th} \) term, which is \( (2(k+1)-1)^2 = (2k+2-1)^2 = (2k+1)^2 \), to both sides:
L.H.S. \( = 1^2 + 3^2 + 5^2 + \dots + (2k-1)^2 + (2k+1)^2 \)
Using P(k), this becomes: \( \frac{k(2k-1)(2k+1)}{3} + (2k+1)^2 \)

\( \implies (2k+1) \left(\frac{k(2k-1)}{3} + (2k+1)\right) \)

\( \implies (2k+1) \left(\frac{k(2k-1) + 3(2k+1)}{3}\right) \)

\( \implies (2k+1) \left(\frac{2k^2-k + 6k+3}{3}\right) \)

\( \implies (2k+1) \left(\frac{2k^2+5k+3}{3}\right) \)
Factor the quadratic term in the parenthesis: \( 2k^2+5k+3 = (2k+3)(k+1) \).

\( \implies \frac{(2k+1)(k+1)(2k+3)}{3} \)
Rearranging the terms for clarity:

\( \implies \frac{(k+1)(2k+1)(2k+3)}{3} \)

\( \implies \frac{(k+1)(2(k+1)-1)(2(k+1)+1)}{3} \)
This is the expression for P(k + 1).
Therefore, P(k + 1) holds whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all natural numbers \( n \in N \).
In simple words: We verified the formula for the first term. Then, we showed that if it holds for 'k', it also works for 'k+1', which means it's true for all natural numbers.

Exam Tip: Pay close attention to factoring quadratic expressions. Make sure you correctly identify the factors that will lead to the \( (k+1) \) version of the formula.

Question 16. Let the given statement be P(n), i.e., \( P(n) : \frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \dots + \frac{1}{(3n-2)(3n+1)} = \frac{n}{3n+1} \).
Answer: Let P(n) represent the given statement:
\( P(n) : \frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \dots + \frac{1}{(3n-2)(3n+1)} = \frac{n}{3n+1} \)
For \( n = 1 \):
L.H.S. \( = \frac{1}{1 \cdot 4} = \frac{1}{4} \)
R.H.S. \( = \frac{1}{3 \cdot 1+1} = \frac{1}{4} \)
Since L.H.S. = R.H.S., P(1) is true.
Assume that P(k) holds for some positive integer k.
So, we have: \( P(k) : \frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \dots + \frac{1}{(3k-2)(3k+1)} = \frac{k}{3k+1} \)
We now aim to demonstrate that P(k + 1) is true, assuming P(k) is true.
The \( (k+1)^{th} \) term is \( \frac{1}{(3(k+1)-2)(3(k+1)+1)} = \frac{1}{(3k+3-2)(3k+3+1)} = \frac{1}{(3k+1)(3k+4)} \).
Add this term to both sides:
L.H.S. \( = \frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \dots + \frac{1}{(3k-2)(3k+1)} + \frac{1}{(3k+1)(3k+4)} \)
Using P(k), this becomes: \( \frac{k}{3k+1} + \frac{1}{(3k+1)(3k+4)} \)

\( \implies \frac{1}{3k+1} \left(k + \frac{1}{3k+4}\right) \)

\( \implies \frac{1}{3k+1} \left(\frac{k(3k+4) + 1}{3k+4}\right) \)

\( \implies \frac{1}{3k+1} \left(\frac{3k^2+4k+1}{3k+4}\right) \)
Factor the numerator: \( 3k^2+4k+1 = (3k+1)(k+1) \).

\( \implies \frac{1}{3k+1} \left(\frac{(3k+1)(k+1)}{3k+4}\right) \)

\( \implies \frac{k+1}{3k+4} \)

\( \implies \frac{k+1}{3(k+1)+1} \)
This is the expression for P(k + 1).
Therefore, P(k + 1) holds whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all natural numbers \( n \in N \).
In simple words: We confirmed the formula for the initial case. Then, we assumed it works for 'k' and added the next term, showing it works for 'k+1', thus proving it for all natural numbers.

Exam Tip: Partial fraction decomposition is often useful for these types of sums, though not strictly required when using induction. Focus on factoring common terms and simplifying algebraic expressions correctly.

Question 17. Let the given statement be P(n), i.e., \( P(n) : \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \frac{1}{7 \cdot 9} + \dots + \frac{1}{(2n+1)(2n+3)} = \frac{n}{3(2n+3)} \).
Answer: Let P(n) represent the given statement:
\( P(n) : \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \frac{1}{7 \cdot 9} + \dots + \frac{1}{(2n+1)(2n+3)} = \frac{n}{3(2n+3)} \)
For \( n = 1 \):
L.H.S. \( = \frac{1}{3 \cdot 5} = \frac{1}{15} \)
R.H.S. \( = \frac{1}{3(2 \cdot 1+3)} = \frac{1}{3(5)} = \frac{1}{15} \)
Since L.H.S. = R.H.S., P(1) is true.
Assume that P(k) holds for some positive integer k.
So, we have: \( P(k) : \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \frac{1}{7 \cdot 9} + \dots + \frac{1}{(2k+1)(2k+3)} = \frac{k}{3(2k+3)} \)
We now aim to demonstrate that P(k + 1) is true, assuming P(k) is true.
The \( (k+1)^{th} \) term is \( \frac{1}{(2(k+1)+1)(2(k+1)+3)} = \frac{1}{(2k+3)(2k+5)} \).
Add this term to both sides:
L.H.S. \( = \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2k+1)(2k+3)} + \frac{1}{(2k+3)(2k+5)} \)
Using P(k), this becomes: \( \frac{k}{3(2k+3)} + \frac{1}{(2k+3)(2k+5)} \)

\( \implies \frac{1}{2k+3} \left(\frac{k}{3} + \frac{1}{2k+5}\right) \)

\( \implies \frac{1}{2k+3} \left(\frac{k(2k+5) + 3}{3(2k+5)}\right) \)

\( \implies \frac{1}{2k+3} \left(\frac{2k^2+5k+3}{3(2k+5)}\right) \)
Factor the numerator: \( 2k^2+5k+3 = (2k+3)(k+1) \).

\( \implies \frac{1}{2k+3} \left(\frac{(2k+3)(k+1)}{3(2k+5)}\right) \)

\( \implies \frac{k+1}{3(2k+5)} \)

\( \implies \frac{k+1}{3(2(k+1)+3)} \)
This is the expression for P(k + 1).
Therefore, P(k + 1) holds whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all natural numbers \( n \in N \).
In simple words: We confirmed the formula for the initial case. Then, we assumed it works for 'k' and added the next term, showing it works for 'k+1', thus proving it for all natural numbers.

Exam Tip: Remember to simplify the terms in the denominators before adding, as this will reveal common factors and make the algebraic simplification much easier. Always check for factorization of quadratic expressions.

Question 18. Prove that \( 1 + 2 + 3 + ... + n < \frac{1}{8} (2n + 1)^2 \) for all natural numbers \( n \).
Answer: Let \( P(n) \) be the statement \( 1 + 2 + 3 + ... + n < \frac{1}{8} (2n + 1)^2 \).
**Base Case (n = 1):**
For \( n = 1 \), the statement becomes \( 1 < \frac{1}{8} (2(1) + 1)^2 \).
This simplifies to \( 1 < \frac{1}{8} (3)^2 \), which is \( 1 < \frac{9}{8} \).
Since \( 1 < 1.125 \), the statement \( P(1) \) is true.
**Inductive Hypothesis:**
Assume that \( P(k) \) is true for some positive integer \( k \).
So, we assume \( 1 + 2 + 3 + ... + k < \frac{1}{8} (2k + 1)^2 \) is true.
**Inductive Step:**
We need to prove that \( P(k+1) \) is true, meaning \( 1 + 2 + 3 + ... + k + (k+1) < \frac{1}{8} (2(k+1) + 1)^2 \).
Starting with the left side of \( P(k+1) \):
\( 1 + 2 + 3 + ... + k + (k+1) \)
From our hypothesis, we know \( 1 + 2 + 3 + ... + k < \frac{1}{8} (2k + 1)^2 \).
So, \( 1 + 2 + 3 + ... + k + (k+1) < \frac{1}{8} (2k + 1)^2 + (k+1) \)
Now, let's simplify the right side:
\( \frac{1}{8} (2k + 1)^2 + (k+1) = \frac{1}{8} (4k^2 + 4k + 1) + (k+1) \)
\( = \frac{4k^2 + 4k + 1 + 8(k+1)}{8} \)
\( = \frac{4k^2 + 4k + 1 + 8k + 8}{8} \)
\( = \frac{4k^2 + 12k + 9}{8} \)
This expression can be rewritten as \( \frac{(2k + 3)^2}{8} \).
Notice that \( 2(k+1) + 1 = 2k + 2 + 1 = 2k + 3 \).
So, \( \frac{(2k + 3)^2}{8} = \frac{(2(k+1) + 1)^2}{8} \).
Therefore, we have shown that \( 1 + 2 + 3 + ... + k + (k+1) < \frac{1}{8} (2k + 3)^2 = \frac{1}{8} [2(k+1) + 1]^2 \).
This proves that \( P(k+1) \) is true.
**Conclusion:**
By the Principle of Mathematical Induction, the statement \( P(n) \) is true for all natural numbers \( n \).
In simple words: We showed that if the statement works for some number, it also works for the next number. And since it works for the first number (n=1), it will work for all numbers.

Exam Tip: For inequalities in mathematical induction, the key is to correctly manipulate the inequality after assuming P(k) and then proving P(k+1) by ensuring the right side is greater than or equal to the target expression.

Question 19. Prove that \( n(n + 1)(n + 5) \) is a multiple of 3 for all natural numbers \( n \).
Answer: Let \( P(n) \) be the statement \( n(n + 1)(n + 5) \) is a multiple of 3.
**Base Case (n = 1):**
For \( n = 1 \), the expression becomes \( 1(1 + 1)(1 + 5) = 1 \cdot 2 \cdot 6 = 12 \).
Since 12 is a multiple of 3 (\( 12 = 3 \cdot 4 \)), the statement \( P(1) \) is true.
**Inductive Hypothesis:**
Assume that \( P(k) \) is true for some positive integer \( k \).
So, we assume \( k(k + 1)(k + 5) \) is a multiple of 3.
This means \( k(k + 1)(k + 5) = 3m \) for some integer \( m \).
Expanding this, \( k(k^2 + 6k + 5) = k^3 + 6k^2 + 5k = 3m \).
**Inductive Step:**
We need to prove that \( P(k+1) \) is true, meaning \( (k+1)((k+1) + 1)((k+1) + 5) \) is a multiple of 3.
This simplifies to \( (k+1)(k+2)(k+6) \).
Let's expand \( (k+1)(k+2)(k+6) \):
\( (k+1)(k^2 + 8k + 12) \)
\( = k(k^2 + 8k + 12) + 1(k^2 + 8k + 12) \)
\( = k^3 + 8k^2 + 12k + k^2 + 8k + 12 \)
\( = k^3 + 9k^2 + 20k + 12 \)
Now, let's rearrange this expression to use our inductive hypothesis \( k^3 + 6k^2 + 5k = 3m \):
\( = (k^3 + 6k^2 + 5k) + (3k^2 + 15k + 12) \)
Substitute \( 3m \) for \( k^3 + 6k^2 + 5k \):
\( = 3m + 3k^2 + 15k + 12 \)
We can factor out 3 from the remaining terms:
\( = 3m + 3(k^2 + 5k + 4) \)
\( = 3(m + k^2 + 5k + 4) \)
Since \( m, k \) are integers, \( m + k^2 + 5k + 4 \) is also an integer. Therefore, \( (k+1)(k+2)(k+6) \) is a multiple of 3.
This proves that \( P(k+1) \) is true.
**Conclusion:**
By the Principle of Mathematical Induction, the statement \( P(n) \) is true for all natural numbers \( n \).
In simple words: We checked if a number times the next number times five more than the next number is always divisible by three. It works for the first number (1), and if it works for any number (k), it also works for the next number (k+1). So, it works for all numbers.

Exam Tip: When proving divisibility, try to rearrange the \( P(k+1) \) expression to explicitly include the \( P(k) \) term (e.g., \( 3m \)) and then show the remaining terms are also divisible by the given number.

Question 20. Prove that \( 10^{2n - 1} + 1 \) is divisible by 11 for every natural number \( n \).
Answer: Let \( P(n) \) be the statement \( 10^{2n - 1} + 1 \) is divisible by 11.
**Base Case (n = 1):**
For \( n = 1 \), the expression becomes \( 10^{2(1) - 1} + 1 = 10^{1} + 1 = 10 + 1 = 11 \).
Since 11 is divisible by 11, the statement \( P(1) \) is true.
**Inductive Hypothesis:**
Assume that \( P(k) \) is true for some positive integer \( k \).
So, we assume \( 10^{2k - 1} + 1 \) is divisible by 11.
This means \( 10^{2k - 1} + 1 = 11m \) for some integer \( m \).
From this, we can write \( 10^{2k - 1} = 11m - 1 \).
**Inductive Step:**
We need to prove that \( P(k+1) \) is true, meaning \( 10^{2(k+1) - 1} + 1 \) is divisible by 11.
Let's simplify the exponent for \( P(k+1) \): \( 2(k+1) - 1 = 2k + 2 - 1 = 2k + 1 \).
So we need to show that \( 10^{2k + 1} + 1 \) is divisible by 11.
Consider the expression:
\( 10^{2k + 1} + 1 = 10^{2k - 1 + 2} + 1 \)
\( = 10^{2} \cdot 10^{2k - 1} + 1 \)
\( = 100 \cdot 10^{2k - 1} + 1 \)
Now substitute \( 10^{2k - 1} = 11m - 1 \) from our inductive hypothesis:
\( = 100(11m - 1) + 1 \)
\( = 100 \cdot 11m - 100 + 1 \)
\( = 1100m - 99 \)
We can factor out 11 from this expression:
\( = 11(100m - 9) \)
Since \( m \) is an integer, \( 100m - 9 \) is also an integer. Therefore, \( 10^{2k + 1} + 1 \) is divisible by 11.
This proves that \( P(k+1) \) is true.
**Conclusion:**
By the Principle of Mathematical Induction, the statement \( P(n) \) is true for all natural numbers \( n \).
In simple words: We proved that the number \( 10 \) raised to a power that is odd, plus \( 1 \), can always be divided by \( 11 \). We showed it works for \( n=1 \), and if it works for \( k \), it also works for \( k+1 \).

Exam Tip: When manipulating terms like \( 10^{2k+1} \), rewrite them to expose the \( 10^{2k-1} \) term from your inductive hypothesis. This helps in direct substitution and simplification.

Question 21. Prove that \( x^{2n} - y^{2n} \) is divisible by \( x + y \) for all natural numbers \( n \).
Answer: Let \( P(n) \) be the statement \( x^{2n} - y^{2n} \) is divisible by \( x + y \).
**Base Case (n = 1):**
For \( n = 1 \), the expression becomes \( x^{2(1)} - y^{2(1)} = x^2 - y^2 \).
We know that \( x^2 - y^2 = (x + y)(x - y) \).
Since \( (x + y)(x - y) \) is clearly divisible by \( x + y \), the statement \( P(1) \) is true.
**Inductive Hypothesis:**
Assume that \( P(k) \) is true for some positive integer \( k \).
So, we assume \( x^{2k} - y^{2k} \) is divisible by \( x + y \).
This means \( x^{2k} - y^{2k} = m(x + y) \) for some integer \( m \).
From this, we can write \( x^{2k} = m(x + y) + y^{2k} \).
**Inductive Step:**
We need to prove that \( P(k+1) \) is true, meaning \( x^{2(k+1)} - y^{2(k+1)} \) is divisible by \( x + y \).
Let's simplify the exponents for \( P(k+1) \): \( 2(k+1) = 2k + 2 \).
So we need to show that \( x^{2k + 2} - y^{2k + 2} \) is divisible by \( x + y \).
Consider the expression:
\( x^{2k + 2} - y^{2k + 2} = x^2 \cdot x^{2k} - y^2 \cdot y^{2k} \)
Now substitute \( x^{2k} = m(x + y) + y^{2k} \) from our inductive hypothesis:
\( = x^2 [m(x + y) + y^{2k}] - y^2 \cdot y^{2k} \)
\( = x^2 m(x + y) + x^2 y^{2k} - y^2 y^{2k} \)
\( = x^2 m(x + y) + y^{2k} (x^2 - y^2) \)
We know \( x^2 - y^2 = (x + y)(x - y) \). Substitute this in:
\( = x^2 m(x + y) + y^{2k} (x + y)(x - y) \)
Now, we can factor out \( (x + y) \) from the entire expression:
\( = (x + y) [x^2 m + y^{2k} (x - y)] \)
Since \( m, x, y, k \) are integers, the term in the square brackets \( [x^2 m + y^{2k} (x - y)] \) is also an integer. Therefore, \( x^{2k + 2} - y^{2k + 2} \) is divisible by \( x + y \).
This proves that \( P(k+1) \) is true.
**Conclusion:**
By the Principle of Mathematical Induction, the statement \( P(n) \) is true for all natural numbers \( n \).
In simple words: We showed that \( x \) to an even power minus \( y \) to the same even power can always be divided by \( x + y \). This holds true for the first number, and if it holds for \( k \), it also holds for \( k+1 \).

Exam Tip: When dealing with divisibility proofs involving two variables like \( x^{2n} - y^{2n} \), the algebraic manipulation (especially factoring and substitution) is crucial for isolating the divisor term \( (x+y) \).

Question 22. Prove that \( 3^{2n+2} - 8n - 9 \) is divisible by 8 for all natural numbers \( n \).
Answer: Let \( P(n) \) be the statement \( 3^{2n+2} - 8n - 9 \) is divisible by 8.
**Base Case (n = 1):**
For \( n = 1 \), the expression becomes \( 3^{2(1)+2} - 8(1) - 9 = 3^4 - 8 - 9 \).
\( = 81 - 8 - 9 = 81 - 17 = 64 \).
Since 64 is divisible by 8 (\( 64 = 8 \cdot 8 \)), the statement \( P(1) \) is true.
**Inductive Hypothesis:**
Assume that \( P(k) \) is true for some positive integer \( k \).
So, we assume \( 3^{2k+2} - 8k - 9 \) is divisible by 8.
This means \( 3^{2k+2} - 8k - 9 = 8m \) for some integer \( m \).
From this, we can write \( 3^{2k+2} = 8m + 8k + 9 \).
**Inductive Step:**
We need to prove that \( P(k+1) \) is true, meaning \( 3^{2(k+1)+2} - 8(k+1) - 9 \) is divisible by 8.
Let's simplify the expression for \( P(k+1) \):
\( 3^{2k+2+2} - 8k - 8 - 9 \)
\( = 3^{2k+4} - 8k - 17 \)
Consider the term \( 3^{2k+4} \):
\( 3^{2k+4} = 3^2 \cdot 3^{2k+2} = 9 \cdot 3^{2k+2} \).
Now substitute \( 3^{2k+2} = 8m + 8k + 9 \) from our inductive hypothesis:
\( = 9(8m + 8k + 9) - 8k - 17 \)
\( = 72m + 72k + 81 - 8k - 17 \)
\( = 72m + (72k - 8k) + (81 - 17) \)
\( = 72m + 64k + 64 \)
We can factor out 8 from this expression:
\( = 8(9m + 8k + 8) \)
Since \( m, k \) are integers, \( 9m + 8k + 8 \) is also an integer. Therefore, \( 3^{2(k+1)+2} - 8(k+1) - 9 \) is divisible by 8.
This proves that \( P(k+1) \) is true.
**Conclusion:**
By the Principle of Mathematical Induction, the statement \( P(n) \) is true for all natural numbers \( n \).
In simple words: We showed that the number \( 3 \) raised to a specific power, minus \( 8 \) times the number, minus \( 9 \), can always be divided by \( 8 \). It holds for the first number, and if it holds for \( k \), it also holds for \( k+1 \).

Exam Tip: Ensure that when you substitute the inductive hypothesis, you expand carefully and group terms to factor out the desired divisor (in this case, 8). Don't lose track of constant terms.

Question 23. Prove that \( 41^n - 14^n \) is a multiple of 27 for all natural numbers \( n \).
Answer: Let \( P(n) \) be the statement \( 41^n - 14^n \) is a multiple of 27.
**Base Case (n = 1):**
For \( n = 1 \), the expression becomes \( 41^1 - 14^1 = 41 - 14 = 27 \).
Since 27 is a multiple of 27 (\( 27 = 27 \cdot 1 \)), the statement \( P(1) \) is true.
**Inductive Hypothesis:**
Assume that \( P(k) \) is true for some positive integer \( k \).
So, we assume \( 41^k - 14^k \) is a multiple of 27.
This means \( 41^k - 14^k = 27m \) for some integer \( m \).
From this, we can write \( 41^k = 27m + 14^k \).
**Inductive Step:**
We need to prove that \( P(k+1) \) is true, meaning \( 41^{k+1} - 14^{k+1} \) is a multiple of 27.
Consider the expression:
\( 41^{k+1} - 14^{k+1} \)
\( = 41 \cdot 41^k - 14 \cdot 14^k \)
Now substitute \( 41^k = 27m + 14^k \) from our inductive hypothesis:
\( = 41(27m + 14^k) - 14 \cdot 14^k \)
\( = 41 \cdot 27m + 41 \cdot 14^k - 14 \cdot 14^k \)
\( = 41 \cdot 27m + (41 - 14) \cdot 14^k \)
\( = 41 \cdot 27m + 27 \cdot 14^k \)
We can factor out 27 from this expression:
\( = 27(41m + 14^k) \)
Since \( m, k \) are integers, \( 41m + 14^k \) is also an integer. Therefore, \( 41^{k+1} - 14^{k+1} \) is a multiple of 27.
This proves that \( P(k+1) \) is true.
**Conclusion:**
By the Principle of Mathematical Induction, the statement \( P(n) \) is true for all natural numbers \( n \).
In simple words: We showed that \( 41 \) to the power of any natural number minus \( 14 \) to the same power always gives a number that can be divided by \( 27 \). It works for the first number and follows the pattern for all subsequent numbers.

Exam Tip: For divisibility proofs, identify the common factor in the \( P(k+1) \) expression after substitution. Often, this involves creating the desired coefficient by splitting terms (e.g., \( 41 \cdot 14^k - 14 \cdot 14^k \)).

Question 24. Prove that \( (2n + 7) < (n + 3)^2 \) for all natural numbers \( n \).
Answer: Let \( P(n) \) be the statement \( (2n + 7) < (n + 3)^2 \).
**Base Case (n = 1):**
For \( n = 1 \), the statement becomes \( (2(1) + 7) < (1 + 3)^2 \).
This simplifies to \( 9 < 4^2 \), which is \( 9 < 16 \).
Since \( 9 < 16 \), the statement \( P(1) \) is true.
**Inductive Hypothesis:**
Assume that \( P(k) \) is true for some positive integer \( k \).
So, we assume \( (2k + 7) < (k + 3)^2 \) is true.
**Inductive Step:**
We need to prove that \( P(k+1) \) is true, meaning \( (2(k+1) + 7) < ((k+1) + 3)^2 \).
This simplifies to \( (2k + 2 + 7) < (k + 4)^2 \), which is \( (2k + 9) < (k + 4)^2 \).
Starting with the left side of the inequality for \( P(k+1) \):
\( 2(k+1) + 7 = 2k + 2 + 7 = 2k + 9 \).
From our inductive hypothesis, we know \( 2k + 7 < (k + 3)^2 \).
Let's add 2 to both sides of the inductive hypothesis:
\( (2k + 7) + 2 < (k + 3)^2 + 2 \)
\( 2k + 9 < k^2 + 6k + 9 + 2 \)
\( 2k + 9 < k^2 + 6k + 11 \).
Now, we need to show that \( k^2 + 6k + 11 < (k + 4)^2 \).
\( (k + 4)^2 = k^2 + 8k + 16 \).
We need to compare \( k^2 + 6k + 11 \) with \( k^2 + 8k + 16 \).
Since \( k \) is a natural number (\( k \ge 1 \)), we know that \( 6k < 8k \) and \( 11 < 16 \).
Specifically, \( k^2 + 8k + 16 - (k^2 + 6k + 11) = 2k + 5 \).
Since \( k \ge 1 \), \( 2k + 5 \) is always positive, so \( k^2 + 6k + 11 < k^2 + 8k + 16 \).
Therefore, \( 2k + 9 < k^2 + 6k + 11 < k^2 + 8k + 16 = (k + 4)^2 \).
This establishes that \( (2(k+1) + 7) < ((k+1) + 3)^2 \), which means \( P(k+1) \) is true.
**Conclusion:**
By the Principle of Mathematical Induction, the statement \( P(n) \) is true for all natural numbers \( n \).
In simple words: We showed that for any natural number, \( 2 \) times that number plus \( 7 \) is always smaller than that number plus \( 3 \), squared. It's true for the first number, and if true for \( k \), it is true for \( k+1 \).

Exam Tip: For inequality proofs by induction, after establishing \( P(k+1) > \text{some expression} \), you often need a secondary inequality to show that "some expression" is less than the desired right-hand side of \( P(k+1) \). Careful comparison of terms is essential.