CBSE Class 11 Mathematics Principle Of Mathematical Induction Worksheet Set 02

Question 1. Let P(n) be the statement \(n(n + 1)\) is an even number then find P(6).
Answer: Given statement is \(P(n) = n(n + 1)\)
To find \(P(6)\), substitute \(n = 6\) into the statement:
\(P(6) = 6(6 + 1)\)
\(P(6) = 6(7) = 42\)
Since 42 is an even number, \(P(6)\) is true.
In simple words: Substitute 6 in place of \(n\) in the given algebraic expression to get the required value.

Exam Tip: Be careful with basic arithmetic substitution; ensure you perform operations inside parentheses first.

Question 2. Prove the following by using the principle of mathematical induction for all \(n \in \mathbb{N}\): \(1.3 + 2.3^2 + 3.3^3 + \ldots + n.3^n = \frac{(2n - 1)3^{n + 1} + 3}{4}\)
Answer: Let the given statement be \(P(n)\): \(1.3 + 2.3^2 + 3.3^3 + \ldots + n.3^n = \frac{(2n - 1)3^{n + 1} + 3}{4}\)
Step 1: Check for \(n = 1\).
LHS = \(1.3^1 = 3\)
RHS = \(\frac{(2(1) - 1)3^{1 + 1} + 3}{4} = \frac{(1)3^2 + 3}{4} = \frac{9 + 3}{4} = \frac{12}{4} = 3\)
Since LHS = RHS, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true for some positive integer \(k\).
\(1.3 + 2.3^2 + 3.3^3 + \ldots + k.3^k = \frac{(2k - 1)3^{k + 1} + 3}{4}\)
Step 3: We must prove that \(P(k + 1)\) is true, i.e.,
\(1.3 + 2.3^2 + \ldots + k.3^k + (k + 1)3^{k + 1} = \frac{(2(k + 1) - 1)3^{(k + 1) + 1} + 3}{4}\)
Starting with LHS of \(P(k + 1)\):
\(= [1.3 + 2.3^2 + \ldots + k.3^k] + (k + 1)3^{k + 1}\)
\(= \frac{(2k - 1)3^{k + 1} + 3}{4} + (k + 1)3^{k + 1}\)
\(= \frac{(2k - 1)3^{k + 1} + 3 + 4(k + 1)3^{k + 1}}{4}\)
\(= \frac{3^{k + 1}[(2k - 1) + 4(k + 1)] + 3}{4}\)
\(= \frac{3^{k + 1}[2k - 1 + 4k + 4] + 3}{4}\)
\(= \frac{3^{k + 1}[6k + 3] + 3}{4}\)
\(= \frac{3^{k + 1} \cdot 3(2k + 1) + 3}{4}\)
\(= \frac{(2k + 1)3^{k + 2} + 3}{4}\)
\(= \frac{(2(k + 1) - 1)3^{(k + 1) + 1} + 3}{4}\) = RHS
Thus, \(P(k + 1)\) is true whenever \(P(k)\) is true. Hence, by mathematical induction, the statement is true for all \(n \in \mathbb{N}\).
In simple words: Establish the base case for \(n=1\), assume it holds for \(n=k\), and algebraically manipulate the expression for \(n=k+1\) using the inductive hypothesis.

Exam Tip: Factor out common powers of terms like \(3^{k+1}\) carefully to simplify the numerator during the inductive step.

Question 3. Prove the following by using the principle of mathematical induction for all \(n \in \mathbb{N}\): \(1 + \frac{1}{(1 + 2)} + \frac{1}{(1 + 2 + 3)} + \ldots + \frac{1}{(1 + 2 + 3 + \ldots + n)} = \frac{2n}{n + 1}\)
Answer: Let \(P(n)\): \(1 + \frac{1}{(1 + 2)} + \frac{1}{(1 + 2 + 3)} + \ldots + \frac{1}{(1 + 2 + 3 + \ldots + n)} = \frac{2n}{n + 1}\)
We know that the sum of first \(n\) natural numbers is \(1 + 2 + 3 + \ldots + n = \frac{n(n + 1)}{2}\).
Thus, the \(n\)-th term can be written as: \(a_n = \frac{1}{\frac{n(n + 1)}{2}} = \frac{2}{n(n + 1)}\)
Step 1: For \(n = 1\), LHS = \(1\), RHS = \(\frac{2(1)}{1 + 1} = \frac{2}{2} = 1\). Thus, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true:
\(1 + \frac{1}{(1 + 2)} + \ldots + \frac{2}{k(k + 1)} = \frac{2k}{k + 1}\)
Step 3: Prove for \(n = k + 1\):
LHS \(= [1 + \frac{1}{(1 + 2)} + \ldots + \frac{2}{k(k + 1)}] + \frac{2}{(k + 1)(k + 2)}\)
\(= \frac{2k}{k + 1} + \frac{2}{(k + 1)(k + 2)}\)
\(= \frac{2}{k + 1} \left( k + \frac{1}{k + 2} \right)\)
\(= \frac{2}{k + 1} \left( \frac{k(k + 2) + 1}{k + 2} \right)\)
\(= \frac{2}{k + 1} \left( \frac{k^2 + 2k + 1}{k + 2} \right)\)
\(= \frac{2(k + 1)^2}{(k + 1)(k + 2)} = \frac{2(k + 1)}{(k + 1) + 1}\) = RHS
Hence, \(P(k + 1)\) is true, and by induction, the statement holds for all \(n \in \mathbb{N}\).
In simple words: Simplify the denominator using the formula for the sum of the first \(n\) natural numbers before applying mathematical induction.

Exam Tip: Rewrite denominators containing arithmetic progressions into compact algebraic fractions before proceeding.

Question 4. Prove by using the principle of mathematical induction \(3^{2n} - 1\) is divisible by 8 for \(n \in \mathbb{N}\).
Answer: Let \(P(n)\): \(3^{2n} - 1\) is divisible by 8.
Step 1: Check for \(n = 1\).
\(3^{2(1)} - 1 = 3^2 - 1 = 9 - 1 = 8\)
Since 8 is divisible by 8, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true for some natural number \(k\), i.e., \(3^{2k} - 1 = 8m\) for some integer \(m\).
Therefore, \(3^{2k} = 8m + 1\).
Step 3: Prove for \(n = k + 1\):
Consider \(3^{2(k + 1)} - 1 = 3^{2k + 2} - 1\)
\(= 3^{2k} \cdot 3^2 - 1\)
\(= 3^{2k} \cdot 9 - 1\)
Substitute \(3^{2k} = 8m + 1\):
\(= (8m + 1)9 - 1\)
\(= 72m + 9 - 1\)
\(= 72m + 8\)
\(= 8(9m + 1)\)
Since \(8(9m + 1)\) is a multiple of 8, it is divisible by 8. Thus, \(P(k + 1)\) is true.
Hence, by the principle of mathematical induction, \(3^{2n} - 1\) is divisible by 8 for all \(n \in \mathbb{N}\).
In simple words: Express the inductive assumption as a multiple of 8, substitute it into the expression for \(k+1\), and factor out 8.

Exam Tip: For divisibility problems, always substitute the expression for the base term (like \(3^{2k}\)) from the induction hypothesis.

Question 5. Prove the following by using the principle of mathematical induction for all \(n \in \mathbb{N}\): \(1 + 3 + 3^2 + \ldots + 3^{n - 1} = \frac{3^n - 1}{2}\)
Answer: Let \(P(n)\): \(1 + 3 + 3^2 + \ldots + 3^{n - 1} = \frac{3^n - 1}{2}\)
Step 1: Check for \(n = 1\).
LHS = \(3^{1 - 1} = 3^0 = 1\)
RHS = \(\frac{3^1 - 1}{2} = \frac{2}{2} = 1\)
Since LHS = RHS, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true for \(n = k\):
\(1 + 3 + 3^2 + \ldots + 3^{k - 1} = \frac{3^k - 1}{2}\)
Step 3: Prove for \(n = k + 1\):
LHS of \(P(k + 1)\) is:
\(= [1 + 3 + 3^2 + \ldots + 3^{k - 1}] + 3^{(k + 1) - 1}\)
\(= \frac{3^k - 1}{2} + 3^k\)
\(= \frac{3^k - 1 + 2 \cdot 3^k}{2}\)
\(= \frac{3 \cdot 3^k - 1}{2}\)
\(= \frac{3^{k + 1} - 1}{2}\) = RHS
Thus, \(P(k + 1)\) is true whenever \(P(k)\) is true. Hence, by induction, the formula holds for all \(n \in \mathbb{N}\).
In simple words: This is the sum of a geometric series; add the \((k+1)\)-th term to the assumed sum and simplify algebraically.

Exam Tip: Recognize geometric progressions to quickly identify the last term for any given \(n\).

Question 6. Prove by using the principle of mathematical induction for all \(n \in \mathbb{N}\). \(1.3 + 3.5 + 5.7 + \ldots + (2n - 1)(2n + 1) = \frac{n(4n^2 + 6n - 1)}{3}\)
Answer: Let \(P(n)\): \(1.3 + 3.5 + 5.7 + \ldots + (2n - 1)(2n + 1) = \frac{n(4n^2 + 6n - 1)}{3}\)
Step 1: For \(n = 1\),
LHS = \((2(1) - 1)(2(1) + 1) = (1)(3) = 3\)
RHS = \(\frac{1(4(1)^2 + 6(1) - 1)}{3} = \frac{4 + 6 - 1}{3} = \frac{9}{3} = 3\)
Since LHS = RHS, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true:
\(1.3 + 3.5 + \ldots + (2k - 1)(2k + 1) = \frac{k(4k^2 + 6k - 1)}{3}\)
Step 3: Prove for \(n = k + 1\):
LHS \(= [1.3 + 3.5 + \ldots + (2k - 1)(2k + 1)] + (2(k + 1) - 1)(2(k + 1) + 1)\)
\(= \frac{k(4k^2 + 6k - 1)}{3} + (2k + 1)(2k + 3)\)
\(= \frac{k(4k^2 + 6k - 1) + 3(2k + 1)(2k + 3)}{3}\)
Expand the numerator:
\(= \frac{4k^3 + 6k^2 - k + 3(4k^2 + 8k + 3)}{3}\)
\(= \frac{4k^3 + 6k^2 - k + 12k^2 + 24k + 9}{3}\)
\(= \frac{4k^3 + 18k^2 + 23k + 9}{3}\)
We need to factor this into \(\frac{(k + 1)(4(k + 1)^2 + 6(k + 1) - 1)}{3}\).
Expanding the target numerator: \((k + 1)(4(k^2 + 2k + 1) + 6k + 6 - 1) = (k + 1)(4k^2 + 8k + 4 + 6k + 5) = (k + 1)(4k^2 + 14k + 9) = 4k^3 + 14k^2 + 9k + 4k^2 + 14k + 9 = 4k^3 + 18k^2 + 23k + 9\).
Thus, LHS = RHS for \(n = k + 1\). Hence, the statement is true for all \(n \in \mathbb{N}\).
In simple words: Substitute \(k+1\), combine terms into a single fraction, expand both numerator and target expression to verify equality.

Exam Tip: When factoring cubic polynomials in induction proofs, expand the target expression from RHS to match coefficients.

Question 7. Prove by using the principle of mathematical induction for all \(n \in \mathbb{N}\): \(1 + 3 + 3^2 + \ldots + 3^{n - 1} = \frac{3^n - 1}{2}\)
Answer: This is identical to Question 5.
Let \(P(n)\): \(1 + 3 + 3^2 + \ldots + 3^{n - 1} = \frac{3^n - 1}{2}\)
Step 1: For \(n = 1\), LHS = \(3^0 = 1\), RHS = \(\frac{3^1 - 1}{2} = 1\). True.
Step 2: Assume for \(n = k\): \(1 + 3 + \ldots + 3^{k - 1} = \frac{3^k - 1}{2}\).
Step 3: For \(n = k + 1\):
LHS \(= \frac{3^k - 1}{2} + 3^k = \frac{3^k - 1 + 2 \cdot 3^k}{2} = \frac{3^{k + 1} - 1}{2}\) = RHS.
Thus, the statement is true for all \(n \in \mathbb{N}\).
In simple words: Combine the assumed sum with the next geometric term to prove the formula for \(k+1\).

Exam Tip: Repeated questions in worksheets should be solved swiftly using standard geometric series formulas.

Question 8. Prove that the product of two consecutive natural numbers is an even number.
Answer: Let the two consecutive natural numbers be \(n\) and \(n + 1\) for any natural number \(n \in \mathbb{N}\).
Their product is \(P = n(n + 1)\).
We consider two cases based on whether \(n\) is even or odd:
Case 1: If \(n\) is even, let \(n = 2k\) for some integer \(k\).
Then \(n(n + 1) = 2k(2k + 1) = 2[k(2k + 1)]\), which is a multiple of 2 and hence even.
Case 2: If \(n\) is odd, let \(n = 2k + 1\) for some integer \(k\).
Then \(n + 1 = 2k + 2 = 2(k + 1)\).
Thus, \(n(n + 1) = (2k + 1) \cdot 2(k + 1) = 2[(2k + 1)(k + 1)]\), which is also a multiple of 2 and hence even.
In all cases, the product of two consecutive natural numbers is even.
In simple words: Any two consecutive numbers always contain one even and one odd number, and their product is always even.

Exam Tip: Case-based analysis on even and odd integers is a robust way to prove divisibility and parity properties.

Question 9. Prove the following by using the principle of mathematical induction for all \(n \in \mathbb{N}\). \((2n + 7) < (n + 3)^2\)
Answer: Let \(P(n)\): \((2n + 7) < (n + 3)^2\)
Step 1: Check for \(n = 1\).
LHS = \(2(1) + 7 = 9\)
RHS = \((1 + 3)^2 = 4^2 = 16\)
Since \(9 < 16\), \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true for some positive integer \(k\):
\((2k + 7) < (k + 3)^2\)
Step 3: Prove for \(n = k + 1\):
We need to prove \((2(k + 1) + 7) < ((k + 1) + 3)^2\), i.e., \((2k + 9) < (k + 4)^2\).
Starting from the inductive hypothesis:
\(2k + 7 < (k + 3)^2\)
Add 2 to both sides:
\(2k + 9 < (k + 3)^2 + 2\)
Notice that \((k + 3)^2 + 2 = k^2 + 6k + 9 + 2 = k^2 + 6k + 11\).
We want to compare this with \((k + 4)^2 = k^2 + 8k + 16\).
Since \(k \ge 1\), \(k^2 + 6k + 11 < k^2 + 8k + 16 = (k + 4)^2\).
Therefore, \((2k + 9) < (k + 4)^2\), which shows that \(P(k + 1)\) is true.
Hence, by mathematical induction, the inequality holds for all \(n \in \mathbb{N}\).
In simple words: Add appropriate terms to both sides of the assumed inequality to build the required inequality for \(k+1\).

Exam Tip: For inequality induction proofs, bridge the gap between \((k+3)^2 + 2\) and \((k+4)^2\) using the fact that \(k \ge 1\).

Question 10. Use the principle of mathematical induction to prove that \(1 + 5 + 9 + 13 + \ldots + (4n - 3) = n(2n - 1)\), \(n \in \mathbb{N}\)
Answer: Let \(P(n)\): \(1 + 5 + 9 + 13 + \ldots + (4n - 3) = n(2n - 1)\)
Step 1: For \(n = 1\), LHS = \(4(1) - 3 = 1\), RHS = \(1(2(1) - 1) = 1\). True.
Step 2: Assume \(P(k)\) is true:
\(1 + 5 + 9 + \ldots + (4k - 3) = k(2k - 1)\)
Step 3: Prove for \(n = k + 1\):
LHS \(= [1 + 5 + \ldots + (4k - 3)] + (4(k + 1) - 3)\)
\(= k(2k - 1) + (4k + 4 - 3)\)
\(= 2k^2 - k + 4k + 1\)
\(= 2k^2 + 3k + 1\)
Factor the quadratic expression:
\(= 2k^2 + 2k + k + 1 = 2k(k + 1) + 1(k + 1) = (2k + 1)(k + 1)\)
\(= (k + 1)(2(k + 1) - 1)\) = RHS
Thus, \(P(k + 1)\) is true. Hence, the statement holds for all \(n \in \mathbb{N}\).
In simple words: Add the \((k+1)\)-th arithmetic term to the assumed sum and factor the resulting quadratic.

Exam Tip: Arithmetic progression sums are straightforward; ensure correct expansion and factoring of the quadratic step.

Question 11. Prove that: \(2.7^n + 3.5^n - 5\) is divisible by 24 for all \(n \in \mathbb{N}\).
Answer: Let \(P(n)\): \(2 \cdot 7^n + 3 \cdot 5^n - 5\) is divisible by 24.
Step 1: Check for \(n = 1\).
\(2 \cdot 7^1 + 3 \cdot 5^1 - 5 = 14 + 15 - 5 = 24\)
Since 24 is divisible by 24, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true for \(n = k\):
\(2 \cdot 7^k + 3 \cdot 5^k - 5 = 24m\) for some integer \(m\).
\(2 \cdot 7^k = 24m - 3 \cdot 5^k + 5\)
Step 3: Prove for \(n = k + 1\):
Consider \(2 \cdot 7^{k + 1} + 3 \cdot 5^{k + 1} - 5 = 2 \cdot 7^k \cdot 7 + 3 \cdot 5^k \cdot 5 - 5\)
Substitute \(2 \cdot 7^k = 24m - 3 \cdot 5^k + 5\):
\(= (24m - 3 \cdot 5^k + 5)7 + 3 \cdot 5^k \cdot 5 - 5\)
\(= 168m - 21 \cdot 5^k + 35 + 15 \cdot 5^k - 5\)
\(= 168m - 6 \cdot 5^k + 30\)
Wait, let us re-group carefully:
\(= 168m - 6(5^k - 5)\)... Let's check with another method (subtraction method).
Consider \(P(k + 1) - 5P(k)\) or similar techniques, or directly analyze \(2 \cdot 7^{k+1} + 3 \cdot 5^{k+1} - 5 - 5(2 \cdot 7^k + 3 \cdot 5^k - 5)\):
\(= 2 \cdot 7^k(7 - 5) + 3 \cdot 5^k(5 - 5)\) - wait, let's use standard substitution:
From \(2 \cdot 7^k = 24m - 3 \cdot 5^k + 5\):
\(2 \cdot 7^{k + 1} + 3 \cdot 5^{k + 1} - 5 = 7(2 \cdot 7^k) + 15 \cdot 5^k - 5\)
\(= 7(24m - 3 \cdot 5^k + 5) + 15 \cdot 5^k - 5\)
\(= 168m - 21 \cdot 5^k + 35 + 15 \cdot 5^k - 5\)
\(= 168m - 6 \cdot 5^k + 30\)
Let's check for \(k=1\): \(168m - 6(5) + 30 = 168m\), which is divisible by 24. Since \(5^k - 5\) is always a multiple of 4 for \(k \ge 1\), \(6(5^k - 5)\) is a multiple of 24.
Thus, the expression is divisible by 24.
In simple words: Substitute the inductive hypothesis into the \(k+1\) expression and group terms into multiples of 24.

Exam Tip: For complex divisibility expressions, express one exponential term in terms of others and simplify.

Question 12. Prove the following by using the principle of mathematical induction for all \(n \in \mathbb{N}\): \(\left(1 + \frac{1}{1}\right)\left(1 + \frac{1}{2}\right)\left(1 + \frac{1}{3}\right) \ldots \left(1 + \frac{1}{n}\right) = (n + 1)\)
Answer: Let \(P(n)\): \(\left(1 + \frac{1}{1}\right)\left(1 + \frac{1}{2}\right)\left(1 + \frac{1}{3}\right) \ldots \left(1 + \frac{1}{n}\right) = (n + 1)\)
Step 1: For \(n = 1\), LHS = \(\left(1 + \frac{1}{1}\right) = 2\), RHS = \(1 + 1 = 2\). True.
Step 2: Assume \(P(k)\) is true:
\(\left(1 + \frac{1}{1}\right)\left(1 + \frac{1}{2}\right) \ldots \left(1 + \frac{1}{k}\right) = k + 1\)
Step 3: Prove for \(n = k + 1\):
LHS \(= \left[ \left(1 + \frac{1}{1}\right)\left(1 + \frac{1}{2}\right) \ldots \left(1 + \frac{1}{k}\right) \right] \left(1 + \frac{1}{k + 1}\right)\)
\(= (k + 1) \left(1 + \frac{1}{k + 1}\right)\)
\(= (k + 1) \left(\frac{k + 1 + 1}{k + 1}\right)\)
\(= (k + 1) \left(\frac{k + 2}{k + 1}\right)\)
\(= k + 2 = (k + 1) + 1\) = RHS
Thus, \(P(k + 1)\) is true. Hence, by induction, the statement holds for all \(n \in \mathbb{N}\).
In simple words: Multiply the assumed product by the \((k+1)\)-th fractional term and simplify.

Answer: Let \(P(n)\): \(3^n < 4^n\)
Step 1: For \(n = 1\), LHS = \(3^1 = 3\), RHS = \(4^1 = 4\). Since \(3 < 4\), \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true for some natural number \(k\):
\(3^k < 4^k\)
Step 3: Prove for \(n = k + 1\):
Multiply both sides of the inequality by 4 (since \(4 > 0\)):
\(4 \cdot 3^k < 4 \cdot 4^k\)
\(4 \cdot 3^k < 4^{k + 1}\)
Since \(3 < 4\), we have \(3 \cdot 3^k < 4 \cdot 3^k\), which implies \(3^{k + 1} < 4 \cdot 3^k\)
Therefore, \(3^{k + 1} < 4 \cdot 3^k < 4^{k + 1}\), so \(3^{k + 1} < 4^{k + 1}\).
Thus, \(P(k + 1)\) is true. Hence, by mathematical induction, \(3^n < 4^n\) for all \(n \in \mathbb{N}\).
In simple words: Multiply the inequality by 4 and use the fact that 3 is smaller than 4 to establish the inequality for \(k+1\).

Exam Tip: Exponential inequalities can be proven easily by multiplying the induction hypothesis by the base of the larger exponential term.

Question 14. For every positive integer \(n\), prove that \(7^n - 3^n\) is divisible by 4.
Answer: Let \(P(n)\): \(7^n - 3^n\) is divisible by 4.
Step 1: For \(n = 1\), \(7^1 - 3^1 = 7 - 3 = 4\). Since 4 is divisible by 4, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true: \(7^k - 3^k = 4m\) for some integer \(m\), so \(7^k = 4m + 3^k\).
Step 3: Prove for \(n = k + 1\):
Consider \(7^{k + 1} - 3^{k + 1} = 7^k \cdot 7 - 3^k \cdot 3\)
Substitute \(7^k = 4m + 3^k\):
\(= (4m + 3^k)7 - 3 \cdot 3^k\)
\(= 28m + 7 \cdot 3^k - 3 \cdot 3^k\)
\(= 28m + 4 \cdot 3^k\)
\(= 4(7m + 3^k)\)
Since \(4(7m + 3^k)\) is a multiple of 4, it is divisible by 4. Thus, \(P(k + 1)\) is true.
Hence, by induction, \(7^n - 3^n\) is divisible by 4 for all positive integers \(n\).
In simple words: Substitute \(7^k = 4m + 3^k\) into the expression for \(k+1\) and factor out 4.

Exam Tip: Standard divisibility proofs follow a strict substitution and factoring pattern.

Question 15. Prove by using the principle of mathematical induction for \(n \in \mathbb{N}\): \((2n + 7) < (n + 3)^2\)
Answer: This is identical to Question 9.
Let \(P(n)\): \((2n + 7) < (n + 3)^2\).
Step 1: For \(n = 1\), \(9 < 16\). True.
Step 2: Assume \(2k + 7 < (k + 3)^2\).
Step 3: For \(n = k + 1\), adding 2 to both sides gives \(2k + 9 < (k + 3)^2 + 2 = k^2 + 6k + 11 < (k + 4)^2\).
Thus, the inequality holds for all \(n \in \mathbb{N}\).
In simple words: Repeated inequality proof verified using algebraic expansion for \(k+1\).

Exam Tip: Ensure inequality steps clearly show that the expanded term is strictly less than the target square.

Question 16. Prove the following by using the principle of mathematical induction for all \(n \in \mathbb{N}\): \(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{2^n} = 1 - \frac{1}{2^n}\)
Answer: Let \(P(n)\): \(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{2^n} = 1 - \frac{1}{2^n}\)
Step 1: For \(n = 1\), LHS = \(\frac{1}{2}\), RHS = \(1 - \frac{1}{2^1} = \frac{1}{2}\). True.
Step 2: Assume \(P(k)\) is true:
\(\frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{2^k} = 1 - \frac{1}{2^k}\)
Step 3: Prove for \(n = k + 1\):
LHS \(= \left[\frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{2^k}\right] + \frac{1}{2^{k + 1}}\)
\(= 1 - \frac{1}{2^k} + \frac{1}{2^{k + 1}}\)
\(= 1 - \left(\frac{2}{2^{k + 1}} - \frac{1}{2^{k + 1}}\right)\)
\(= 1 - \frac{1}{2^{k + 1}}\) = RHS
Thus, \(P(k + 1)\) is true. Hence, the formula holds for all \(n \in \mathbb{N}\).
In simple words: Add the next power of two to the assumed sum to show that the resulting fraction simplifies correctly.

Exam Tip: Geometric sums with fractions resolve quickly by combining common denominators.

Question 17. Suppose \(P(n)\): \(n(n + 1)(n + 2)\) is divisible by 6. Show that \(P(1)\), \(P(2)\) and \(P(3)\) are true.
Answer: Given statement: \(P(n) = n(n + 1)(n + 2)\)
- For \(n = 1\): \(P(1) = 1(1 + 1)(1 + 2) = 1(2)(3) = 6\), which is divisible by 6. True.
- For \(n = 2\): \(P(2) = 2(2 + 1)(2 + 2) = 2(3)(4) = 24\), which is divisible by 6. True.
- For \(n = 3\): \(P(3) = 3(3 + 1)(3 + 2) = 3(4)(5) = 60\), which is divisible by 6. True.
In simple words: Substitute 1, 2, and 3 into the expression and check if the resulting products are multiples of 6.

Exam Tip: Direct substitution questions require simple arithmetic evaluation.

Question 18. Let \(P(n)\) be the statement \(n^2 > 25\), prove that whenever \(P(k)\) is true, \(P(k + 1)\) is also true.
Answer: Assume that \(P(k)\) is true for some integer \(k\), i.e., \(k^2 > 25\).
We need to show that \(P(k + 1)\) is true, i.e., \((k + 1)^2 > 25\).
Since \(k^2 > 25\), taking square roots gives \(k > 5\) (considering positive domain).
Adding 1 to both sides: \(k + 1 > 6\)
Squaring both sides: \((k + 1)^2 > 6^2 = 36\)
Since \(36 > 25\), it follows that \((k + 1)^2 > 25\).
Thus, whenever \(P(k)\) is true, \(P(k + 1)\) is also true.
In simple words: Use the inductive assumption to bound \(k+1\) and show its square exceeds 25.

Answer: Given statement: \(P(n) = n^2 - n + 41\)
- For \(n = 1\): \(P(1) = 1^2 - 1 + 41 = 41\) (Prime number). True.
- For \(n = 2\): \(P(2) = 2^2 - 2 + 41 = 43\) (Prime number). True.
- For \(n = 3\): \(P(3) = 3^2 - 3 + 41 = 47\) (Prime number). True.
- For \(n = 41\): \(P(41) = 41^2 - 41 + 41 = 41^2 = 1681\)
Since \(1681 = 41 \times 41\), it is not a prime number. Thus, \(P(41)\) is false.
In simple words: Evaluating polynomial expressions for small values can yield primes, but higher values can fail primality tests.

Exam Tip: Always test polynomial prime generators at the critical value matching the constant term (like 41).

Question 20. Explain the principle of mathematical induction.
Answer: The principle of mathematical induction is a mathematical proof technique used to prove that a statement \(P(n)\) is true for all natural numbers \(n \in \mathbb{N}\). It consists of two essential steps:
1. Base Step: Verify that the statement is true for the initial value \(n = 1\) (i.e., prove \(P(1)\) is true).
2. Inductive Step: Assume that the statement is true for some arbitrary natural number \(n = k\) (Inductive Hypothesis, i.e., assume \(P(k)\) is true), and using this assumption, prove that the statement is also true for \(n = k + 1\) (i.e., prove \(P(k + 1)\) is true).
Once both steps are completed, the statement is proven true for all natural numbers \(n\).
In simple words: Induction works like falling dominoes: if the first domino falls, and whenever any domino falls it knocks over the next one, then all dominoes will fall.

Exam Tip: Clearly state both the base case and the inductive step when asked to define or explain mathematical induction.