CBSE Class 11 Mathematics Principle Of Mathematical Induction Worksheet Set 04

CBSE Class 11 Mathematics Worksheet - Principle of Mathematical Induction (3). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question. By principle of mathematical induction show that \( 3^{2n+2} - 8n - 9 \) is divisible by 8 for all \( n \in N \).
Answer: Let \( P(n): 3^{2n+2} - 8n - 9 \) is divisible by 8.

(i) Let \( P(1): 3^{2(1)+2} - 8(1) - 9 = 3^4 - 17 = 81 - 17 = 64 \), which is divisible by 8.
\( \therefore P(1) \) is true.

(ii) Let \( P(k) \) be true:
\( P(k): 3^{2k+2} - 8k - 9 = 8m \) where \( m \in N \)
\( \Rightarrow 3^{2k+2} = 8m + 8k + 9 \)

(iii) To prove \( P(k+1) \) is true:
\( P(k+1): 3^{2(k+1)+2} - 8(k+1) - 9 \) is divisible by 8.
L.H.S. \( = 3^{2k+4} - 8k - 8 - 9 \)
\( = 3^{2k+2} \cdot 3^2 - 8k - 17 \)
\( = [8m + 8k + 9] \cdot 9 - 8k - 17 \qquad \{\text{from } P(k)\} \)
\( = 72m + 72k + 81 - 8k - 17 \)
\( = 72m + 64k + 64 \)
\( = 8(9m + 8k + 8) \), which is divisible by 8.

\( \therefore P(k+1) \) is true.
\( \therefore \) By the principle of Mathematical Induction, \( P(n) \) is true for all \( n \in N \).

Question. Prove by PMI, \( 3^{2n} \) when divided by 8, the remainder is always 1.
Answer: Let \( P(n): 3^{2n} \) when divided by 8 leaves remainder 1.

(i) Let \( P(1): 3^{2(1)} = 9 = 8(1) + 1 \)
Clearly, \( P(1) \) is true.

(ii) Let \( P(k) \) be true:
\( P(k): 3^{2k} = 8m + 1 \qquad \{m \in N\} \)

(iii) To prove \( P(k+1) \) is true:
\( P(k+1): 3^{2(k+1)} \)
\( = 3^{2k} \cdot 3^2 \)
\( = (8m + 1) \cdot 9 \qquad \{\text{from } P(k)\} \)
\( = 72m + 9 \)
\( = 8(9m + 1) + 1 \), which clearly leaves remainder 1 when divided by 8.

\( \therefore P(k+1) \) is true.
\( \therefore \) By PMI, \( P(n) \) is true for all \( n \in N \).

Question. By PMI, show \( x^{2n} - y^{2n} \) is divisible by \( x + y \).
Answer: Let \( P(n): x^{2n} - y^{2n} \) is divisible by \( x + y \).

(i) For \( n = 1 \):
\( P(1): x^2 - y^2 = (x + y)(x - y) \), which is clearly divisible by \( x + y \).
\( \therefore P(1) \) is true.

(ii) Let \( P(k) \) be true:
\( P(k): x^{2k} - y^{2k} = (x + y)m \qquad \{m \in N\} \)
\( \Rightarrow x^{2k} = (x + y)m + y^{2k} \qquad \dots (1) \)

(iii) To prove \( P(k+1) \) is true:
\( P(k+1): x^{2k+2} - y^{2k+2} \) is divisible by \( x + y \).
L.H.S. \( = x^{2k} \cdot x^2 - y^{2k} \cdot y^2 \)
\( = [(x + y)m + y^{2k}]x^2 - y^{2k} \cdot y^2 \qquad \{\text{from } (1)\} \)
\( = (x + y)mx^2 + y^{2k} \cdot x^2 - y^{2k} \cdot y^2 \)
\( = (x + y)mx^2 + y^{2k}(x^2 - y^2) \)
\( = (x + y)mx^2 + y^{2k}(x + y)(x - y) \)
\( = (x + y)[mx^2 + y^{2k}(x - y)] \), which is clearly divisible by \( x + y \).

\( \therefore P(k+1) \) is true.
\( \therefore \) By PMI, \( P(n) \) is true for all \( n \in N \).

Question. Show by PMI, \( 11^{n+2} + 12^{2n+1} \) is multiple of 133.
Answer: Let \( P(n): 11^{n+2} + 12^{2n+1} \) is a multiple of 133.

(i) For \( n = 1 \):
\( P(1): 11^{1+2} + 12^{2(1)+1} = 11^3 + 12^3 = 1331 + 1728 = 3059 \)
Since \( 3059 = 23 \times 133 \), it is divisible by 133.
Clearly, \( P(1) \) is true.

(ii) Let \( P(k) \) be true:
\( P(k): 11^{k+2} + 12^{2k+1} = 133m \qquad \{m \in N\} \)
\( \Rightarrow 11^{k+2} = 133m - 12^{2k+1} \)

(iii) To prove \( P(k+1) \) is true:
\( P(k+1): 11^{k+3} + 12^{2k+3} \) is a multiple of 133.
L.H.S. \( = 11^{k+2} \cdot 11 + 12^{2k+1} \cdot 12^2 \)
\( = (133m - 12^{2k+1}) \cdot 11 + 12^{2k+1} \cdot 144 \)
\( = 133m \times 11 - 11 \cdot 12^{2k+1} + 144 \cdot 12^{2k+1} \)
\( = 133m \times 11 + 12^{2k+1}(144 - 11) \)
\( = 133m \times 11 + 12^{2k+1} \cdot 133 \br /> \( = 133[11m + 12^{2k+1}] \), which is divisible by 133.

\( \therefore P(k+1) \) is true.
\( \therefore \) By PMI, \( P(n) \) is true for all \( n \in N \).

Question. By PMI, show that \( 2 \cdot 7^n + 3 \cdot 5^n - 5 \) is divisible by 24.
Answer: Let \( P(n): 2 \cdot 7^n + 3 \cdot 5^n - 5 \) is divisible by 24.

(i) For \( n = 1 \):
\( P(1): 2 \cdot 7^1 + 3 \cdot 5^1 - 5 = 14 + 15 - 5 = 24 \), which is divisible by 24.
\( \therefore P(1) \) is true.

(ii) Let \( P(k) \) be true:
\( P(k): 2 \cdot 7^k + 3 \cdot 5^k - 5 = 24m \qquad \{m \in N\} \)
\( \Rightarrow 2 \cdot 7^k = 24m - 3 \cdot 5^k + 5 \)

(iii) To prove \( P(k+1) \) is true:
\( P(k+1): 2 \cdot 7^{k+1} + 3 \cdot 5^{k+1} - 5 \) is divisible by 24.
L.H.S. \( = 2 \cdot 7^k \cdot 7 + 3 \cdot 5^{k+1} - 5 \)
\( = (24m - 3 \cdot 5^k + 5) \cdot 7 + 3 \cdot 5^{k+1} - 5 \)
\( = 24m \times 7 - 21 \cdot 5^k + 35 + 15 \cdot 5^k - 5 \)
\( = 24m \times 7 - 6 \cdot 5^k + 30 \)
\( = 24m \times 7 - 6(5^k - 5) \)

Now, \( 5^k - 5 \) is always a multiple of 4 for all values of \( k \in N \):
For \( k = 1 \): \( 5^1 - 5 = 0 = 4 \times 0 \)
For \( k = 2 \): \( 5^2 - 5 = 25 - 5 = 20 = 4 \times 5 \)
For \( k = 3 \): \( 5^3 - 5 = 125 - 5 = 120 = 4 \times 30 \), and so on.
Therefore, we can write \( 5^k - 5 = 4p \) where \( p \in N \).
\( \Rightarrow \text{L.H.S.} = 24m \times 7 - 6(4p) = 24(7m - p) \), which is divisible by 24.

\( \dots P(k+1) \) is true.
\( \therefore \) By PMI, \( P(n) \) is true for all \( n \in N \).

Question. Show by PMI, \( n(n+1)(n+5) \) is multiple of 3.
Answer: Let \( P(n): n(n+1)(n+5) \) is a multiple of 3.

(i) For \( n = 1 \):
\( P(1) = 1(1+1)(1+5) = (2)(6) = 12 \), which is a multiple of 3.
\( \therefore P(1) \) is true.

(ii) Let \( P(k) \) be true:
\( P(k): k(k+1)(k+5) = 3m \qquad \{m \in N\} \)
\( \Rightarrow k(k^2 + 6k + 5) = 3m \)
\( \Rightarrow k^3 + 6k^2 + 5k = 3m \qquad \dots (1) \)

(iii) To prove \( P(k+1) \) is true:
\( P(k+1): (k+1)(k+2)(k+6) \) is a multiple of 3.
L.H.S. \( = (k+1)(k^2 + 8k + 12) \)
\( = k^3 + 8k^2 + 12k + k^2 + 8k + 12 \)
\( = k^3 + 9k^2 + 20k + 12 \)
\( = (k^3 + 6k^2 + 5k) + (3k^2 + 15k + 12) \)
\( = 3m + 3k^2 + 15k + 12 \qquad \{\text{from } (1)\} \)
\( = 3(m + k^2 + 5k + 4) \), which is a multiple of 3.

\( \therefore P(k+1) \) is true.
\( \therefore \) By PMI, \( P(n) \) is true for all \( n \in N \).

Question. Prove by induction that the sum of the cubes of three consecutive natural numbers is divisible by 9.
Answer: Let the three consecutive natural numbers be \( n, (n + 1), (n + 2) \).
Let \( P(n): n^3 + (n+1)^3 + (n+2)^3 \) is divisible by 9.

(i) For \( n = 1 \):
\( P(1): 1^3 + (1+1)^3 + (1+2)^3 = 1 + 8 + 27 = 36 \), which is divisible by 9.
\( \therefore P(1) \) is true.

(ii) Let \( P(k) \) be true:
\( P(k): k^3 + (k+1)^3 + (k+2)^3 = 9m \qquad \{m \in N\} \)
\( \Rightarrow 3k^3 + 9k^2 + 15k + 9 = 9m \qquad \dots (1) \)

(iii) To prove \( P(k+1) \) is true:
\( P(k+1): (k+1)^3 + (k+2)^3 + (k+3)^3 \) is divisible by 9.
L.H.S. \( = (k+1)^3 + (k+2)^3 + (k^3 + 9k^2 + 27k + 27) \)
\( = k^3 + 3k^2 + 3k + 1 + k^3 + 8k^2 + 12k + 8 + k^3 + 9k^2 + 27k + 27 \)
\( = 3k^3 + 18k^2 + 42k + 36 \)
\( = (3k^3 + 9k^2 + 15k + 9) + (9k^2 + 27k + 27) \)
\( = 9m + 9(k^2 + 3k + 3) \qquad \{\text{from } (1)\} \)
\( = 9(m + k^2 + 3k + 3) \), which is divisible by 9.

\( \therefore P(k+1) \) is true.
\( \therefore \) By PMI, \( P(n) \) is true for all \( n \in N \).