OP Malhotra Class 11 Maths Solutions Chapter 8 Mathematical Induction Exercise 8 (B)

Using the Principle of Mathematical Induction Prove That:

Question 1. \( n (n + 1) (n + 5) \) is a multiple of 6.
Answer: Let \( P (n) \) be the statement: \( n (n + 1) (n + 5) \) is a multiple of 6.
For \( n = 1 \): \( 1 (1 + 1) (1 + 5) = 1 \times 2 \times 6 = 12 \). Since 12 is a multiple of 6, \( P (1) \) is true.
Assume that \( P (m) \) is true for some natural number \( m \).
This means \( m (m + 1) (m + 5) \) is a multiple of 6.
So, \( m (m + 1) (m + 5) = 6k \) for some integer \( k \in N \) ...(1)
Now, we need to prove that \( P (m + 1) \) is true.
\( P (m + 1) \): \( (m + 1) ((m + 1) + 1) ((m + 1) + 5) \)
\( = (m + 1) (m + 2) (m + 6) \)
\( = (m + 1) (m^2 + 8m + 12) \)
\( = (m + 1) (m^2 + 5m + 3m + 12) \)
\( = (m + 1) (m^2 + 5m) + (m + 1) (3m + 12) \)
\( = m (m + 1) (m + 5) + 3m (m + 1) + 12 (m + 1) \)
From (1), we know \( m (m + 1) (m + 5) = 6k \). We also know that the product of three consecutive integers is always divisible by 6.
\( = 6k + 3(2m)(m+1) + 12(m+1) \) This step shows how common factors like 3 and 2 are grouped for divisibility.
The product \( m(m+1) \) is always even, so \( 3m(m+1) \) is a multiple of 6. Let \( 3m(m+1) = 6k' \).
\( = 6k + 6k' + 12 (m + 1) \)
\( = 6 [k + k' + 2 (m + 1)] \)
Let \( k'' = k + k' + 2 (m + 1) \). Since \( k, k', m \) are integers, \( k'' \) is also an integer and \( k'' \in N \).
So, \( (m + 1) (m + 2) (m + 6) = 6k'' \), which means it is a multiple of 6.
Thus, \( P (m + 1) \) is true.
Therefore, by the principle of mathematical induction, \( P (n) \) is true for all natural numbers \( n \).
In simple words: We start by checking if the statement works for \( n=1 \). Then, we assume it works for some number \( m \). Finally, we use that assumption to prove it also works for \( m+1 \). This proves it works for all natural numbers.

🎯 Exam Tip: Remember to clearly state the base case \( P(1) \), the inductive hypothesis \( P(m) \), and the inductive step \( P(m+1) \), showing how the multiple of 6 (or any divisor) is maintained at each step.

Question 2. \( n^3 + (n + 1)^3 + (n + 2)^3 \) is a multiple of 9.
Answer: Let \( P (n) \) be the statement: \( n^3 + (n + 1)^3 + (n + 2)^3 \) is a multiple of 9.
For \( n = 1 \): \( 1^3 + (1 + 1)^3 + (1 + 2)^3 \)
\( = 1^3 + 2^3 + 3^3 \)
\( = 1 + 8 + 27 \)
\( = 36 \). Since 36 is a multiple of 9, \( P (1) \) is true.
Assume that \( P (m) \) is true for some natural number \( m \).
This means \( m^3 + (m + 1)^3 + (m + 2)^3 \) is a multiple of 9.
So, \( m^3 + (m + 1)^3 + (m + 2)^3 = 9k \) for some integer \( k \in N \) ...(1)
Now, we need to prove that \( P (m + 1) \) is true.
\( P (m + 1) \): \( (m + 1)^3 + ((m + 1) + 1)^3 + ((m + 1) + 2)^3 \)
\( = (m + 1)^3 + (m + 2)^3 + (m + 3)^3 \)
We can rearrange this expression to use our assumption from (1):
\( = (m + 1)^3 + (m + 2)^3 + (m^3 + 9m^2 + 27m + 27) - m^3 \). This step helps to isolate the original \( P(m) \) expression.
\( = (m^3 + (m + 1)^3 + (m + 2)^3) - m^3 + (m + 3)^3 \)
Using equation (1) and expanding \( (m+3)^3 = m^3 + 3 \cdot m^2 \cdot 3 + 3 \cdot m \cdot 3^2 + 3^3 = m^3 + 9m^2 + 27m + 27 \):
\( = 9k - m^3 + (m^3 + 9m^2 + 27m + 27) \)
\( = 9k + 9m^2 + 27m + 27 \)
\( = 9 (k + m^2 + 3m + 3) \)
Let \( \lambda = k + m^2 + 3m + 3 \). Since \( k, m \) are integers, \( \lambda \) is also an integer and \( \lambda \in N \).
So, \( (m + 1)^3 + (m + 2)^3 + (m + 3)^3 = 9\lambda \), which means it is a multiple of 9.
Thus, \( P (m + 1) \) is true.
Therefore, by the principle of mathematical induction, \( P (n) \) is true for all natural numbers \( n \).
In simple words: This problem shows that the sum of the cubes of any three numbers in a row can always be divided by 9. We check this for the first set of three numbers, then assume it works for any 'm' and show it must also work for 'm+1'.

🎯 Exam Tip: When proving divisibility with induction, expanding terms carefully and looking for opportunities to substitute the inductive hypothesis (e.g., \( 9k \) or \( 6k \)) are key steps. Rearranging terms creatively is often needed.

Question 3. \( 2^{3n}-1 \) is divisible by 7.
Answer: Let \( P (n) \) be the statement: \( 2^{3n} - 1 \) is divisible by 7.
For \( n = 1 \): \( 2^{3 \times 1} - 1 \)
\( = 2^3 - 1 \)
\( = 8 - 1 = 7 \). Since 7 is divisible by 7, \( P (1) \) is true.
Assume that \( P (m) \) is true for some natural number \( m \).
This means \( 2^{3m} - 1 \) is divisible by 7.
So, \( 2^{3m} - 1 = 7k \) for some integer \( k \in N \) ...(1)
From this, we can write \( 2^{3m} = 7k + 1 \). This substitution will be useful later.
Now, we need to prove that \( P (m + 1) \) is true.
\( P (m + 1) \): \( 2^{3(m+1)} - 1 \)
\( = 2^{3m + 3} - 1 \)
\( = 2^{3m} \cdot 2^3 - 1 \)
\( = 2^{3m} \cdot 8 - 1 \)
Substitute \( 2^{3m} = 7k + 1 \) from (1):
\( = (7k + 1) \cdot 8 - 1 \)
\( = 56k + 8 - 1 \)
\( = 56k + 7 \)
\( = 7 (8k + 1) \)
Let \( k' = 8k + 1 \). Since \( k \) is an integer, \( k' \) is also an integer and \( k' \in N \).
So, \( 2^{3(m+1)} - 1 = 7k' \), which means it is divisible by 7.
Thus, \( P (m + 1) \) is true.
Therefore, by the principle of mathematical induction, \( P (n) \) is true for all natural numbers \( n \).
In simple words: We check if \( 2^{3n}-1 \) can be divided by 7 for the first number, then assume it works for 'm', and show it will also work for 'm+1' using algebra. This proves it always works.

🎯 Exam Tip: When you have a term like \( 2^{3m} \), remember to use the inductive hypothesis to express it (e.g., \( 2^{3m} = 7k+1 \)) before substituting into the \( P(m+1) \) expression. This is a common and important step.

Question 4. \( 3^{2n} - 1 \) is divisible by 8.
Answer: Let \( P (n) \) be the statement: \( 3^{2n} - 1 \) is divisible by 8.
For \( n = 1 \): \( 3^{2 \times 1} - 1 \)
\( = 3^2 - 1 \)
\( = 9 - 1 = 8 \). Since 8 is divisible by 8, \( P (1) \) is true.
Assume that \( P (m) \) is true for some natural number \( m \).
This means \( 3^{2m} - 1 \) is divisible by 8.
So, \( 3^{2m} - 1 = 8k \) for some integer \( k \in N \) ...(1)
From this, we can write \( 3^{2m} = 8k + 1 \). This helps in the next step.
Now, we need to prove that \( P (m + 1) \) is true.
\( P (m + 1) \): \( 3^{2(m+1)} - 1 \)
\( = 3^{2m + 2} - 1 \)
\( = 3^{2m} \cdot 3^2 - 1 \)
\( = 3^{2m} \cdot 9 - 1 \)
Substitute \( 3^{2m} = 8k + 1 \) from (1):
\( = (8k + 1) \cdot 9 - 1 \)
\( = 72k + 9 - 1 \)
\( = 72k + 8 \)
\( = 8 (9k + 1) \)
Let \( k' = 9k + 1 \). Since \( k \) is an integer, \( k' \) is also an integer and \( k' \in N \).
So, \( 3^{2(m+1)} - 1 = 8k' \), which means it is divisible by 8.
Thus, \( P (m + 1) \) is true.
Therefore, by the principle of mathematical induction, \( P (n) \) is true for all natural numbers \( n \).
In simple words: This shows that if you take 3, raise it to the power of \( 2n \) (an even number), and then subtract 1, the result will always be perfectly divisible by 8. It's a pattern that holds true for all natural numbers.

🎯 Exam Tip: When the exponent contains a multiple of n (like \( 2n \) or \( 3n \)), remember to expand \( 3^{2(m+1)} \) as \( 3^{2m} \cdot 3^2 \) to easily substitute the inductive hypothesis term \( 3^{2m} \).

Question 5. \( 4^n + 15n - 1 \) is divisible by 9.
Answer: Let \( P (n) \) be the statement: \( 4^n + 15n - 1 \) is divisible by 9.
For \( n = 1 \): \( 4^1 + 15(1) - 1 \)
\( = 4 + 15 - 1 \)
\( = 18 \). Since 18 is divisible by 9, \( P (1) \) is true.
Assume that \( P (m) \) is true for some natural number \( m \).
This means \( 4^m + 15m - 1 \) is divisible by 9.
So, \( 4^m + 15m - 1 = 9k \) for some integer \( k \in N \) ...(1)
From this, we can write \( 4^m = 9k - 15m + 1 \). This substitution is key.
Now, we need to prove that \( P (m + 1) \) is true.
\( P (m + 1) \): \( 4^{m+1} + 15(m + 1) - 1 \)
\( = 4^m \cdot 4 + 15m + 15 - 1 \)
\( = 4^m \cdot 4 + 15m + 14 \)
Substitute \( 4^m = 9k - 15m + 1 \) from (1):
\( = (9k - 15m + 1) \cdot 4 + 15m + 14 \)
\( = 36k - 60m + 4 + 15m + 14 \)
\( = 36k - 45m + 18 \)
\( = 9 (4k - 5m + 2) \)
Let \( k' = 4k - 5m + 2 \). Since \( k, m \) are integers, \( k' \) is also an integer and \( k' \in N \).
So, \( 4^{m+1} + 15(m + 1) - 1 = 9k' \), which means it is divisible by 9.
Thus, \( P (m + 1) \) is true.
Therefore, by the principle of mathematical induction, \( P (n) \) is true for all natural numbers \( n \).
In simple words: We are proving that for any natural number \( n \), the value of \( 4^n + 15n - 1 \) can always be divided by 9 without any remainder. This is done by checking the first case and then using the 'm' to 'm+1' step.

🎯 Exam Tip: For induction problems involving sums or mixed terms, isolate the \( P(m) \) expression or its components (like \( 4^m \)) from the inductive hypothesis to substitute them cleanly into the \( P(m+1) \) expression. Careful algebraic manipulation is crucial.

Question 6. \( 3^{4n+2} + 5^{2n+1} \) is a multiple of 14.
Answer: Let \( P (n) \) be the statement: \( 3^{4n+2} + 5^{2n+1} \) is a multiple of 14.
For \( n = 1 \): \( 3^{4(1)+2} + 5^{2(1)+1} \)
\( = 3^6 + 5^3 \)
\( = 729 + 125 \)
\( = 854 \). To check if 854 is a multiple of 14, we divide 854 by 14, which gives 61.
Since 854 is clearly a multiple of 14, \( P (1) \) is true.
Assume that \( P (m) \) is true for some natural number \( m \).
This means \( 3^{4m+2} + 5^{2m+1} \) is a multiple of 14.
So, \( 3^{4m+2} + 5^{2m+1} = 14k \) for some integer \( k \in N \) ...(1)
From this, we can write \( 3^{4m+2} = 14k - 5^{2m+1} \). This substitution is helpful.
Now, we need to prove that \( P (m + 1) \) is true.
\( P (m + 1) \): \( 3^{4(m+1)+2} + 5^{2(m+1)+1} \)
\( = 3^{4m+4+2} + 5^{2m+2+1} \)
\( = 3^{4m+2} \cdot 3^4 + 5^{2m+1} \cdot 5^2 \)
\( = 3^{4m+2} \cdot 81 + 5^{2m+1} \cdot 25 \)
Substitute \( 3^{4m+2} = 14k - 5^{2m+1} \) from (1):
\( = (14k - 5^{2m+1}) \cdot 81 + 5^{2m+1} \cdot 25 \)
\( = 14k \cdot 81 - 5^{2m+1} \cdot 81 + 5^{2m+1} \cdot 25 \)
\( = 14k \cdot 81 + 5^{2m+1} (25 - 81) \)
\( = 14k \cdot 81 + 5^{2m+1} (-56) \)
\( = 14k \cdot 81 - 5^{2m+1} \cdot 56 \)
\( = 14 (81k - 5^{2m+1} \cdot 4) \)
Let \( k' = 81k - 4 \cdot 5^{2m+1} \). Since \( k, m \) are integers, \( k' \) is also an integer and \( k' \in N \).
So, \( 3^{4(m+1)+2} + 5^{2(m+1)+1} = 14k' \), which means it is a multiple of 14.
Thus, \( P (m + 1) \) is true.
Therefore, by the principle of mathematical induction, \( P (n) \) is true for all natural numbers \( n \).
In simple words: This problem shows that a specific combination of powers of 3 and 5 will always be a multiple of 14 for any natural number \( n \). We follow the standard induction steps to prove this mathematical pattern.

🎯 Exam Tip: When substituting a complex term like \( 3^{4m+2} \), be very careful with the algebraic manipulation, especially when factoring out the common divisor (here, 14). Double-check the signs and coefficients in the final step.

Question 7. \( 7^{2n} + (2^{3n-3}) 3^{n-1} \) is divisible by 25.
Answer: Let \( P (n) \) be the statement: \( 7^{2n} + 2^{3n-3} \cdot 3^{n-1} \) is divisible by 25.
For \( n = 1 \): \( 7^{2(1)} + 2^{3(1)-3} \cdot 3^{1-1} \)
\( = 7^2 + 2^0 \cdot 3^0 \)
\( = 49 + 1 \cdot 1 \)
\( = 49 + 1 = 50 \). Since 50 is clearly divisible by 25, \( P (1) \) is true.
Assume that \( P (m) \) is true for some natural number \( m \).
This means \( 7^{2m} + 2^{3m-3} \cdot 3^{m-1} \) is divisible by 25.
So, \( 7^{2m} + 2^{3m-3} \cdot 3^{m-1} = 25k \) for some integer \( k \in N \) ...(1)
From this, we can write \( 2^{3m-3} \cdot 3^{m-1} = 25k - 7^{2m} \). This will be useful for substitution.
Now, we need to prove that \( P (m + 1) \) is true.
\( P (m + 1) \): \( 7^{2(m+1)} + 2^{3(m+1)-3} \cdot 3^{(m+1)-1} \)
\( = 7^{2m+2} + 2^{3m+3-3} \cdot 3^m \)
\( = 7^{2m} \cdot 7^2 + 2^{3m} \cdot 3^m \)
\( = 7^{2m} \cdot 49 + 2^{3m} \cdot 3^m \)
We can also write \( 2^{3m} \cdot 3^m = (2^3)^m \cdot 3^m = 8^m \cdot 3^m = (24)^m \).
And \( 2^{3m-3} \cdot 3^{m-1} = 2^{3m} \cdot 2^{-3} \cdot 3^m \cdot 3^{-1} = (2^{3m} \cdot 3^m) / (2^3 \cdot 3^1) = (24)^m / 24 \).
So, \( (24)^m = 24 \cdot (25k - 7^{2m}) \) from the adjusted (1) for \( 2^{3m-3} \cdot 3^{m-1} \). This helps to link the terms directly.
\( = 7^{2m} \cdot 49 + 24 \cdot (25k - 7^{2m}) \)
\( = 49 \cdot 7^{2m} + 24 \cdot 25k - 24 \cdot 7^{2m} \)
\( = 25 \cdot 7^{2m} + 24 \cdot 25k \)
\( = 25 (7^{2m} + 24k) \)
Let \( k' = 7^{2m} + 24k \). Since \( k, m \) are integers, \( k' \) is also an integer and \( k' \in N \).
So, \( P (m + 1) \) is divisible by 25.
Thus, \( P (m + 1) \) is true.
Therefore, by the principle of mathematical induction, \( P (n) \) is true for all natural numbers \( n \).
In simple words: This statement proves that a sum involving powers of 7, 2, and 3 will always be perfectly divisible by 25 for any whole number \( n \). We verify the first case and then use algebraic substitution to prove the pattern continues.

🎯 Exam Tip: When terms involve different bases and powers, simplify them into a common base or isolate the part that matches your inductive hypothesis as \( 2^{3m-3} \cdot 3^{m-1} \) or similar, to make substitution easier.

Question 8. \( x^n - y^n \) is divisible by \( x - y \) when \( n \) is an even positive integer.
Answer: Let \( P (n) \) be the statement: \( x^n - y^n \) is divisible by \( x - y \).
For \( n = 1 \): \( x^1 - y^1 = x - y \). This is clearly divisible by \( x - y \). So, \( P (1) \) is true.
Assume that \( P (m) \) is true for some natural number \( m \).
This means \( x^m - y^m \) is divisible by \( x - y \).
So, \( x^m - y^m = (x - y) f(x, y) \) for some polynomial function \( f(x, y) \) in \( x \) and \( y \) ...(1)
Now, we need to prove that \( P (m + 1) \) is true.
\( P (m + 1) \): \( x^{m+1} - y^{m+1} \)
We can rewrite this expression to include terms from the inductive hypothesis:
\( = x^{m+1} - x^m y + x^m y - y^{m+1} \)
\( = x^m (x - y) + y (x^m - y^m) \)
Substitute \( x^m - y^m = (x - y) f(x, y) \) from (1):
\( = x^m (x - y) + y (x - y) f(x, y) \)
Now, we can factor out \( (x - y) \):
\( = (x - y) [x^m + y f(x, y)] \)
Let \( Q(x, y) = x^m + y f(x, y) \). Since \( f(x,y) \) is a polynomial, \( Q(x,y) \) is also a polynomial.
So, \( x^{m+1} - y^{m+1} = (x - y) Q(x, y) \), which means it is clearly divisible by \( x - y \).
Thus, \( P (m + 1) \) is true.
Therefore, by the principle of mathematical induction, \( P (n) \) is true for all natural numbers \( n \).
*Note: The question states "when n is an even positive integer." The proof above holds for all natural numbers \( n \), which means it certainly holds for even positive integers as well. The condition "even" might be relevant for a different proof method or if \( x=y \), but for divisibility by \( x-y \), it is true for all \( n \). A key algebraic property is that \( x^n - y^n \) always has a factor of \( (x-y) \).*
In simple words: This rule means that if you subtract \( y^n \) from \( x^n \), the answer will always be perfectly divisible by \( x-y \). This is true for any whole number \( n \), including even numbers.

🎯 Exam Tip: For divisibility proofs like this, a common algebraic trick is to add and subtract a term (e.g., \( x^m y \)) to create groups that can be factored, allowing you to use the inductive hypothesis effectively. This method is very powerful for algebraic induction.

Question 9. Prove by the method of mathematical induction that \( 3^{2n-1} + 3^n + 4 \) is divisible by 2 for all \( n \in N \).
Answer: Let \( P (n) \) be the statement: \( 3^{2n-1} + 3^n + 4 \) is divisible by 2.
For \( n = 1 \): \( 3^{2(1)-1} + 3^1 + 4 \)
\( = 3^1 + 3 + 4 \)
\( = 3 + 3 + 4 = 10 \). Since 10 is divisible by 2, \( P (1) \) is true.
Assume that \( P (m) \) is true for some natural number \( m \).
This means \( 3^{2m-1} + 3^m + 4 \) is divisible by 2.
So, \( 3^{2m-1} + 3^m + 4 = 2k \) for some integer \( k \in N \) ...(1)
From this, we can write \( 3^{2m-1} = 2k - 3^m - 4 \). This substitution is crucial.
Now, we need to prove that \( P (m + 1) \) is true.
\( P (m + 1) \): \( 3^{2(m+1)-1} + 3^{m+1} + 4 \)
\( = 3^{2m+2-1} + 3^{m+1} + 4 \)
\( = 3^{2m+1} + 3^{m+1} + 4 \)
\( = 3^{2m-1} \cdot 3^2 + 3^m \cdot 3 + 4 \)
\( = (2k - 3^m - 4) \cdot 9 + 3^m \cdot 3 + 4 \)
\( = 18k - 9 \cdot 3^m - 36 + 3 \cdot 3^m + 4 \)
\( = 18k - 6 \cdot 3^m - 32 \)
\( = 2 (9k - 3 \cdot 3^m - 16) \)
Let \( k' = 9k - 3 \cdot 3^m - 16 \). Since \( k, m \) are integers, \( k' \) is also an integer and \( k' \in N \).
So, \( P (m + 1) \) is divisible by 2.
Thus, \( P (m + 1) \) is true.
Therefore, by the principle of mathematical induction, \( P (n) \) is true for all natural numbers \( n \).
In simple words: This problem asks us to prove that for any natural number \( n \), the sum of \( 3^{2n-1} \), \( 3^n \), and 4 will always be an even number, meaning it can be divided by 2. We use mathematical induction to show this.

🎯 Exam Tip: Pay close attention to the exponents when breaking down terms (e.g., \( 3^{2m+1} \) into \( 3^{2m-1} \cdot 3^2 \)). It's vital to create a term that matches the one you can substitute from your inductive hypothesis.

Question 10. \( n < 2^n \) for all \( n \in N \).
Answer: Let \( P (n) \) be the statement: \( n < 2^n \).
For \( n = 1 \): \( 1 < 2^1 \). This simplifies to \( 1 < 2 \), which is true. So, \( P (1) \) is true.
Assume that \( P (m) \) is true for some natural number \( m \).
This means \( m < 2^m \) ...(1)
Now, we need to prove that \( P (m + 1) \) is true.
\( P (m + 1) \): We want to show \( (m + 1) < 2^{m+1} \).
From (1), we know \( m < 2^m \).
Let's consider \( m + 1 \). We know \( 1 \leq 2^m \) for all natural numbers \( m \). In fact, for \( m \geq 1 \), \( 2^m \) is always greater than or equal to 2.
We can add 1 to both sides of the inductive hypothesis \( m < 2^m \):
\( m + 1 < 2^m + 1 \)
We also know that \( 1 < 2^m \) for all \( m \in N \). This is because \( 2^1 = 2 \), \( 2^2 = 4 \), etc., which are all greater than 1.
So, \( 2^m + 1 < 2^m + 2^m \)
\( 2^m + 1 < 2 \cdot 2^m \)
\( 2^m + 1 < 2^{m+1} \)
Combining the inequalities:
\( m + 1 < 2^m + 1 < 2^{m+1} \)
Therefore, \( m + 1 < 2^{m+1} \).
Thus, \( P (m + 1) \) is true.
Hence, by the principle of mathematical induction, \( P (n) \) is true for all natural numbers \( n \).
In simple words: This proof shows that for any counting number \( n \), the number itself will always be smaller than 2 raised to the power of that number. For instance, 3 is less than \( 2^3=8 \).

🎯 Exam Tip: When proving inequalities by induction, start by using the inductive hypothesis directly. Then, try to manipulate the inequality to reach \( P(m+1) \), often by adding or multiplying terms, making sure the direction of the inequality is preserved. Showing \( 2^m + 1 < 2^{m+1} \) is a common step here.

Question 11. \( (2n + 7) < (n + 3)^2 \) for all natural numbers.
Answer: Let \( P (n) \) be the statement: \( (2n + 7) < (n + 3)^2 \).
For \( n = 1 \): \( 2(1) + 7 < (1 + 3)^2 \)
\( 2 + 7 < (4)^2 \)
\( 9 < 16 \). This is true. So, \( P (1) \) is true.
Assume that \( P (m) \) is true for some natural number \( m \).
This means \( (2m + 7) < (m + 3)^2 \) ...(1)
Now, we need to prove that \( P (m + 1) \) is true.
\( P (m + 1) \): We want to show \( (2(m + 1) + 7) < ((m + 1) + 3)^2 \).
This simplifies to \( (2m + 2 + 7) < (m + 4)^2 \), or \( (2m + 9) < (m + 4)^2 \).
From (1), we have \( 2m + 7 < (m + 3)^2 \).
Let's consider the left side of \( P(m+1) \): \( 2m + 9 \).
We know \( 2m + 9 = (2m + 7) + 2 \).
Using (1), we can say: \( (2m + 7) + 2 < (m + 3)^2 + 2 \). This means \( 2m + 9 < m^2 + 6m + 9 + 2 \).
So, \( 2m + 9 < m^2 + 6m + 11 \).
Now, let's look at the right side of \( P(m+1) \): \( (m + 4)^2 \).
\( (m + 4)^2 = m^2 + 8m + 16 \).
We need to show that \( m^2 + 6m + 11 < m^2 + 8m + 16 \).
To do this, we can subtract \( m^2 + 6m + 11 \) from both sides to see if the difference is positive:
\( (m^2 + 8m + 16) - (m^2 + 6m + 11) \)
\( = m^2 - m^2 + 8m - 6m + 16 - 11 \)
\( = 2m + 5 \).
Since \( m \) is a natural number, \( m \geq 1 \). Therefore, \( 2m + 5 \) is always a positive number \( (2(1) + 5 = 7 > 0) \).
This means \( m^2 + 6m + 11 < m^2 + 8m + 16 \).
So, we have \( (2m + 9) < (m^2 + 6m + 11) < (m^2 + 8m + 16) \).
Therefore, \( (2m + 9) < (m + 4)^2 \).
Thus, \( P (m + 1) \) is true.
Hence, by the principle of mathematical induction, \( P (n) \) is true for all natural numbers \( n \).
In simple words: This mathematical statement means that if you take twice a number \( n \) and add 7, the result will always be smaller than what you get when you add 3 to \( n \) and then square that sum. We prove this true for all counting numbers using induction.

🎯 Exam Tip: When proving inequalities, after using the inductive hypothesis for the \( P(m+1) \) left side, you often need a separate step to show that your intermediate expression is indeed less than the \( P(m+1) \) right side. This might involve subtracting expressions or comparing terms directly.

Question 12. \( 1^2 + 2^2 + 3^2 + ... + n^2 > \frac{n^3}{3} \) for all \( n \in N \).
Answer: Let \( P (n) \) be the statement: \( 1^2 + 2^2 + 3^2 + ... + n^2 > \frac{n^3}{3} \).
For \( n = 1 \): \( 1^2 > \frac{1^3}{3} \)
\( 1 > \frac{1}{3} \). This is true. So, \( P (1) \) is true.
Assume that \( P (m) \) is true for some natural number \( m \).
This means \( 1^2 + 2^2 + 3^2 + ... + m^2 > \frac{m^3}{3} \) ...(1)
Now, we need to prove that \( P (m + 1) \) is true.
\( P (m + 1) \): We want to show \( 1^2 + 2^2 + 3^2 + ... + m^2 + (m + 1)^2 > \frac{(m + 1)^3}{3} \).
From (1), we can write:
\( 1^2 + 2^2 + 3^2 + ... + m^2 + (m + 1)^2 > \frac{m^3}{3} + (m + 1)^2 \)
Now, we need to show that \( \frac{m^3}{3} + (m + 1)^2 > \frac{(m + 1)^3}{3} \).
Let's expand and simplify the inequality:
\( \frac{m^3}{3} + m^2 + 2m + 1 > \frac{m^3 + 3m^2 + 3m + 1}{3} \)
Multiply by 3 to clear the denominators:
\( m^3 + 3m^2 + 6m + 3 > m^3 + 3m^2 + 3m + 1 \)
Subtract \( m^3 + 3m^2 \) from both sides:
\( 6m + 3 > 3m + 1 \)
Subtract \( 3m \) from both sides:
\( 3m + 3 > 1 \)
Subtract 3 from both sides:
\( 3m > -2 \)
Since \( m \) is a natural number, \( m \geq 1 \). So \( 3m \) will always be greater than -2. This inequality is true for all \( m \in N \).
Since \( \frac{m^3}{3} + (m + 1)^2 > \frac{(m + 1)^3}{3} \) is true, and we know \( 1^2 + 2^2 + ... + m^2 + (m + 1)^2 > \frac{m^3}{3} + (m + 1)^2 \),
Therefore, \( 1^2 + 2^2 + 3^2 + ... + (m + 1)^2 > \frac{(m + 1)^3}{3} \).
Thus, \( P (m + 1) \) is true.
Hence, by the principle of mathematical induction, \( P (n) \) is true for all natural numbers \( n \).
In simple words: This problem shows that if you sum up the squares of numbers from 1 up to \( n \), the total will always be bigger than \( n \) cubed divided by 3. We prove this using induction, checking the first case and then showing the pattern continues.

🎯 Exam Tip: When proving inequalities with sums, after applying the inductive hypothesis, you often need to prove a smaller inequality (e.g., \( \frac{m^3}{3} + (m + 1)^2 > \frac{(m + 1)^3}{3} \)) by algebraic manipulation. Make sure each step in the algebraic proof is sound and preserves the inequality direction.