CBSE Class 11 Mathematics Principle Of Mathematical Induction Worksheet Set 03

CBSE Class 11 Mathematics Worksheet - Principle of Mathematical Induction (2). 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. Show by Induction, \( \frac{1}{2.5} + \frac{1}{5.8} + \dots + \frac{1}{(3n-1)(3n+2)} = \frac{n}{6n+4} \)
Answer: Let \( P(n): \frac{1}{2.5} + \frac{1}{5.8} + \dots + \frac{1}{(3n-1)(3n+2)} = \frac{n}{6n+4} \)

(i) For \( n = 1 \):
\( P(1): \frac{1}{2.5} = \frac{1}{6(1)+4} \Rightarrow \frac{1}{10} = \frac{1}{10} \)
\( \therefore P(1) \) is true.

(ii) Let \( P(k) \) be true:
\( P(k): \frac{1}{2.5} + \frac{1}{5.8} + \dots + \frac{1}{(3k-1)(3k+2)} = \frac{k}{6k+4} \)

(iii) To prove \( P(k+1) \) is true:
\( P(k+1): \frac{1}{2.5} + \frac{1}{5.8} + \dots + \frac{1}{(3k-1)(3k+2)} + \frac{1}{(3k+2)(3k+5)} = \frac{k+1}{6k+10} \)

Taking L.H.S.:
\( \frac{1}{2.5} + \frac{1}{5.8} + \dots + \frac{1}{(3k-1)(3k+2)} + \frac{1}{(3k+2)(3k+5)} \)
\( = \frac{k}{6k+4} + \frac{1}{(3k+2)(3k+5)} \qquad \text{ \{from } P(k)\} \)
\( = \frac{k}{2(3k+2)} + \frac{1}{(3k+2)(3k+5)} \)
\( = \frac{k(3k+5) + 2}{2(3k+2)(3k+5)} \)
\( = \frac{3k^2 + 5k + 2}{2(3k+2)(3k+5)} \)
\( = \frac{3k^2 + 3k + 2k + 2}{2(3k+2)(3k+5)} \)
\( = \frac{3k(k+1) + 2(k+1)}{2(3k+2)(3k+5)} \)
\( = \frac{(k+1)(3k+2)}{2(3k+2)(3k+5)} \)
\( = \frac{k+1}{2(3k+5)} = \frac{k+1}{6k+10} = \text{R.H.S.} \)

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

Question. Show for all \( n \in N \), \( \frac{1}{1.2.3} + \frac{1}{2.3.4} + \dots + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)} \)
Answer: Let \( P(n): \frac{1}{1.2.3} + \frac{1}{2.3.4} + \dots + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)} \)

(i) For \( n = 1 \):
\( P(1): \frac{1}{1.2.3} = \frac{1(1+3)}{4(1+1)(1+2)} \Rightarrow \frac{1}{6} = \frac{4}{4(2)(3)} = \frac{1}{6} \)
Clearly \( P(1) \) is true.

(ii) Let \( P(k) \) be true:
\( P(k): \frac{1}{1.2.3} + \frac{1}{2.3.4} + \dots + \frac{1}{k(k+1)(k+2)} = \frac{k(k+3)}{4(k+1)(k+2)} \)

(iii) To prove \( P(k+1) \) is true:
\( P(k+1): \frac{1}{1.2.3} + \frac{1}{2.3.4} + \dots + \frac{1}{k(k+1)(k+2)} + \frac{1}{(k+1)(k+2)(k+3)} = \frac{(k+1)(k+4)}{4(k+2)(k+3)} \)

Taking L.H.S.:
\( \frac{1}{1.2.3} + \frac{1}{2.3.4} + \dots + \frac{1}{k(k+1)(k+2)} + \frac{1}{(k+1)(k+2)(k+3)} \)
\( = \frac{k(k+3)}{4(k+1)(k+2)} + \frac{1}{(k+1)(k+2)(k+3)} \qquad \text{ \{from } P(k)\} \)
\( = \frac{k(k+3)^2 + 4}{4(k+1)(k+2)(k+3)} \)
\( = \frac{k(k^2 + 6k + 9) + 4}{4(k+1)(k+2)(k+3)} \)
\( = \frac{k^3 + 6k^2 + 9k + 4}{4(k+1)(k+2)(k+3)} \)
\( = \frac{(k+1)(k^2 + 5k + 4)}{4(k+1)(k+2)(k+3)} \)
\( = \frac{(k+1)(k+4)(k+1)}{4(k+1)(k+2)(k+3)} \)
\( = \frac{(k+1)(k+4)}{4(k+2)(k+3)} = \text{R.H.S.} \)

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

Question. Show using PMI, \( 1.3 + 3.5 + \dots + (2n-1)(2n+1) = \frac{n(4n^2+6n-1)}{3} \)
Answer: Let \( P(n): 1.3 + 3.5 + \dots + (2n-1)(2n+1) = \frac{n(4n^2+6n-1)}{3} \)

(i) For \( n = 1 \):
\( P(1): 1.3 = \frac{1(4(1)^2+6(1)-1)}{3} \Rightarrow 3 = \frac{9}{3} = 3 \)
\( \therefore P(1) \) is true.

(ii) Let \( P(k) \) be true:
\( P(k): 1.3 + 3.5 + \dots + (2k-1)(2k+1) = \frac{k(4k^2+6k-1)}{3} \)

(iii) To prove \( P(k+1) \) is true:
\( P(k+1): 1.3 + 3.5 + \dots + (2k-1)(2k+1) + (2k+1)(2k+3) = \frac{(k+1)[4(k+1)^2+6(k+1)-1]}{3} \)

Taking L.H.S.:
\( 1.3 + 3.5 + \dots + (2k-1)(2k+1) + (2k+1)(2k+3) \)
\( = \frac{k(4k^2+6k-1)}{3} + (2k+1)(2k+3) \qquad \text{ \{from } P(k)\} \)
\( = \frac{4k^3 + 6k^2 - k + 3(4k^2 + 8k + 3)}{3} \)
\( = \frac{4k^3 + 6k^2 - k + 12k^2 + 24k + 9}{3} \)
\( = \frac{4k^3 + 18k^2 + 23k + 9}{3} \)
\( = \frac{(k+1)(4k^2 + 14k + 9)}{3} \)
\( = \frac{(k+1)[4(k^2 + 2k + 1) + 6k + 6 - 1]}{3} \)
\( = \frac{(k+1)[4(k+1)^2 + 6(k+1) - 1]}{3} = \text{R.H.S.} \)

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

Question. Show by Induction \( a + ar + ar^2 + \dots + ar^{n-1} = a \left( \frac{r^n - 1}{r - 1} \right) \)
Answer: Let \( P(n): a + ar + ar^2 + \dots + ar^{n-1} = a \left( \frac{r^n - 1}{r - 1} \right) \)

(i) For \( n = 1 \):
\( P(1): a = a \left( \frac{r^1-1}{r-1} \right) \Rightarrow a = a \)
\( \therefore P(1) \) is true.

(ii) Let \( P(k) \) be true:
\( P(k): a + ar + ar^2 + \dots + ar^{k-1} = a \left( \frac{r^k - 1}{r - 1} \right) \)

(iii) To prove \( P(k+1) \) is true:
\( P(k+1): a + ar + ar^2 + \dots + ar^{k-1} + ar^k = a \left( \frac{r^{k+1} - 1}{r - 1} \right) \)

Taking L.H.S.:
\( a + ar + ar^2 + \dots + ar^{k-1} + ar^k \)
\( = a \left( \frac{r^k - 1}{r - 1} \right) + ar^k \qquad \text{ \{from } P(k)\} \)
\( = a \left[ \frac{r^k - 1 + r^k(r-1)}{r - 1} \right] \)
\( = a \left[ \frac{r^k - 1 + r^{k+1} - r^k}{r - 1} \right] \)
\( = a \left[ \frac{r^{k+1} - 1}{r - 1} \right] = \text{R.H.S.} \)

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

Question. Show by Induction \( 1.3 + 2.3^2 + \dots + n.3^n = \frac{(2n-1)3^{n+1}+3}{4} \)
Answer: Let \( P(n): 1.3 + 2.3^2 + \dots + n.3^n = \frac{(2n-1)3^{n+1}+3}{4} \)

(i) For \( n = 1 \):
\( P(1): 1.3 = \frac{(2(1)-1)3^{1+1}+3}{4} \Rightarrow 3 = \frac{1(9)+3}{4} = 3 \)
\( \therefore P(1) \) is true.

(ii) Let \( P(k) \) be true:
\( P(k): 1.3 + 2.3^2 + \dots + k.3^k = \frac{(2k-1)3^{k+1}+3}{4} \)

(iii) To prove \( P(k+1) \) is true:
\( P(k+1): 1.3 + 2.3^2 + \dots + k.3^k + (k+1)3^{k+1} = \frac{(2(k+1)-1)3^{k+2}+3}{4} = \frac{(2k+1)3^{k+2}+3}{4} \)

Taking L.H.S.:
\( 1.3 + 2.3^2 + \dots + k.3^k + (k+1)3^{k+1} \)
\( = \frac{(2k-1)3^{k+1}+3}{4} + (k+1)3^{k+1} \qquad \text{ \{from } P(k)\} \)
\( = \frac{(2k-1)3^{k+1} + 3 + 4(k+1)3^{k+1}}{4} \)
\( = \frac{3^{k+1}[ (2k-1) + 4k + 4 ] + 3}{4} \)
\( = \frac{3^{k+1}(6k + 3) + 3}{4} \)
\( = \frac{3^{k+1} \cdot 3(2k+1) + 3}{4} \)
\( = \frac{(2k+1)3^{k+2}+3}{4} = \text{R.H.S.} \)

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