Class 11: Mathematical Induction – Exercise 12.2 (Part II) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 26: $2.7^n+3.5^n-5$ is divisible by $24$ for all $n \in N$

Answer:

Let $P(n) :$ $2.7^n+3.5^n-5$ is divisible by $24$ for all $n \in N$

For $n = 1$ , L.H.S $= 2 \times 7 + 3 \times 5 - 5 = 24$ which is divisible by $24$

Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$2.7^k+3.5^k-5$ is divisible by $24$ for all $k \in N$

$\therefore 2.7^k+3.5^k-5 = 24 \lambda$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$\Rightarrow 2.7^{k+1}+3.5^{k+1}-5$

$= 7( 2.7^k) + 5( 3.5^k) - 5$

$= 7 ( 24 \lambda - 3.5^k + 5) + 5 ( 3.5^k) - 5$

$= 7 \times 24 \lambda - 7 ( 3.5^k) + 35 + 5 ( 3.5^k) - 5$

$= 7 \times 24 \lambda - 2 ( 3.5^k) + 30$

$= 7 \times 24 \lambda - 6 ( 5^k - 5)$

We know that $5^k - 5$ is divisible by $4$ . Hence

$= 7 \times 24 \lambda - 24 \mu$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$2.7^n+3.5^n-5$ is divisible by $24$ for all $n \in N$

$\\$

Question 27: $11^{n+2}+ 12^{2n+1}$ is divisible by $133$ for all $n \in N$

Answer:

Let $P(n) :$ $11^{n+2}+ 12^{2n+1}$ is divisible by $133$ for all $n \in N$

For $n = 1$ , L.H.S $= 11^3 + 12^3 = 3059 = 133 \times 23$ which is divisible by $133$

Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$11^{k+2}+ 12^{2k+1}$ is divisible by $133$ for all $k \in N$

$\therefore 11^{k+2}+ 12^{2k+1} = 133 \lambda$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$\Rightarrow 11^{(k+1)+2}+ 12^{2(k+1)+1}$

$= 11. 11^{k+2} + 144 .12^{2k+1}$

$= 11( 133 \lambda - 12^{2k+1}) + 144 . 12^{2k+1}$

$= 11 \times 133 \lambda - 11. 12^{2k+1} + 144 . 12^{2k+1}$

$= 11 \times 133 \lambda + 133. 12^{2k+1}$

$= 133 ( 11 \lambda + 12 ^{2k+1})$

$= 133 \mu$ where $\mu = 11 \lambda + 12 ^{2k+1}$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$2.7^n+3.5^n-5$ is divisible by $24$ for all $n \in N$

$\\$

Question 28: $1\times 1! + 2\times 2! +3\times 3! + \ldots + n\times n! = (n+1)! - 1$ for all $n \in N$

Answer:

Let $P(n) :$ $1\times 1! + 2\times 2! +3\times 3! + \ldots + n\times n! = (n+1)! - 1$ for all $n \in N$

For $n = 1$ , L.H.S $= 1\times 1! = 1$ R.H.S $= 2! - 1 = 1$ Therefore LHS = RHS

Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$1\times 1! + 2\times 2! +3\times 3! + \ldots + k\times k! = (k+1)! - 1$ for all $k \in N$ … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$\Rightarrow 1\times 1! + 2\times 2! +3\times 3! + \ldots + k\times k! + (k+1)\times (k+1)!$

$= (k+1)! - 1 + (k+1) \times (k+1)!$

$= (k+1)! [ 1 + (k+1)] - 1$

$= (k+1)! \times (k+2) - 1$

$= (k+2)! - 1$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$1\times 1! + 2\times 2! +3\times 3! + \ldots + n\times n! = (n+1)! - 1$ for all $n \in N$

$\\$

Question 29: $n^3 - 7n+3$ is divisible by $3$ for all $n \in N$

Answer:

Let $P(n) :$ $n^3 - 7n+3$ is divisible by $3$ for all $n \in N$

For $n = 1$ , L.H.S $= 1 - 7 + 3 = -3$ which is divisible by $3$

Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$k^3 - 7k+3$ is divisible by $3$ for all $k \in N$

$\therefore k^3 - 7k+3 = 3 \lambda$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$\Rightarrow (k+1)^3 - 7(k+1)+3$

$= k^3 + 1 + 3k^2 + 3k - 7 k - 7 + 3$

$= (k^3 - 7 k + 3) + ( 3k^2 + 3k - 6)$

$= 3 \lambda + 3 ( k^2 + k - 2)$

$= 3 \lambda + 3 \mu$ where $\mu = k^2 + k - 2$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$n^3 - 7n+3$ is divisible by $3$ for all $n \in N$

$\\$

Question 30: $1 + 2 + 2^2 + \ldots + 2^n = 2^{n+1} - 1$ for all $n \in N$

Answer:

Let $P(n) :$ $1 + 2 + 2^2 + \ldots + 2^n = 2^{n+1} - 1$ for all $n \in N$

For $n = 1$ , L.H.S $= 1+ 2^1 = 3$ R.H.S $= 2^2 - 1 = 3$ Therefore LHS = RHS

Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$1 + 2 + 2^2 + \ldots + 2^k = 2^{k+1} - 1$ for all $k \in N$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$\Rightarrow 1 + 2 + 2^2 + \ldots + 2^k + 2^{k+1}$

$= 2^{k+1} - 1 + 2^{k+1}$

$= 2. 2^{k+1} - 1$

$= 2^{k+2}-1$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$1 + 2 + 2^2 + \ldots + 2^n = 2^{n+1} - 1$ for all $n \in N$

$\\$

Question 31: $\text{ Prove that} \ 7 + 77 + 777+ \ldots + \underbrace{777 \ldots \ldots 7 }_{n-digits} =\frac{7}{81} (10^{n+1} - 9n-10) \text{ for all } n \in N$

Answer:

Let $P(n) :$ $\text{ Prove that} \ 7 + 77 + 777+ \ldots + \underbrace{777 \ldots \ldots 7 }_{n-digits} =\frac{7}{81} (10^{n+1} - 9n-10) \text{ for all } n \in N$

For $n = 7$ , L.H.S $= 7$ R.H.S $= \frac{7}{81} ( 10^2 - 9 - 10) = 7$ Therefore LHS = RHS

Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$\text{ Prove that} \ 7 + 77 + 777+ \ldots + \underbrace{777 \ldots \ldots 7 }_{k-digits} =\frac{7}{81} (10^{k+1} - 9k-10) \text{ for all } k \in N$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$\Rightarrow 7 + 77 + 777+ \ldots + \underbrace{777 \ldots \ldots 7 }_{k-digits} + \underbrace{777 \ldots \ldots 7 }_{k+1-digits}$

$=\frac{7}{81} (10^{k+1} - 9k-10) + \underbrace{777 \ldots \ldots 7 }_{k+1-digits}$

$=\frac{7}{81} (10^{k+1} - 9k-10) + 7 ( \underbrace{111 \ldots \ldots 1 }_{k+1-digits} )$

$=\frac{7}{81} (10^{k+1} - 9k-10) + \frac{7}{9} ( \underbrace{999 \ldots \ldots 9 }_{k+1-digits} )$

$=\frac{7}{81} (10^{k+1} - 9k-10) + \frac{7}{9} ( 10^{k+1} - 1)$

$=\frac{7}{81} (10^{k+1} - 9k-10) + \frac{7}{81} ( 9.10^{k+1} - 9)$

$= \frac{7}{81} ( 10^{k+1} - 9k - 10 + 9.10^{k+1} - 9)$

$= \frac{7}{81} ( 10^{k+2} - 9 ( k+1) - 10)$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$\text{ Prove that} \ 7 + 77 + 777+ \ldots + \underbrace{777 \ldots \ldots 7 }_{n-digits} =\frac{7}{81} (10^{n+1} - 9n-10) \text{ for all } n \in N$

$\\$

Question 32: Prove that $\frac{n^7}{7}$ $+$ $\frac{n^5}{5}$ $+$ $\frac{n^3}{3}$ $+$ $\frac{n^2}{2}$ $-$ $\frac{37}{210}$ $n$ is a positive integer got all $n \in N$

Answer:

Let $P(n) :$ $\frac{n^7}{7}$ $+$ $\frac{n^5}{5}$ $+$ $\frac{n^3}{3}$ $+$ $\frac{n^2}{2}$ $-$ $\frac{37}{210}$ $n$ is a positive integer got all $n \in N$

For $n = 1$ , L.H.S $= \frac{1}{7} + \frac{1}{5} + \frac{1}{3} + \frac{1}{2} - \frac{37}{210} = \frac{30+42+70+105-37}{210} = \frac{210}{210}$ $= 1$ which is positive.

Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$\frac{k^7}{7}$ $+$ $\frac{k^5}{5}$ $+$ $\frac{k^3}{3}$ $+$ $\frac{k^2}{2}$ $-$ $\frac{37}{210}$ $k$ is a positive integer got all $k \in N$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$\Rightarrow$ $\frac{(k+1)^7}{7} + \frac{(k+1)^5}{5} + \frac{(k+1)^3}{3} + \frac{(k+1)^2}{2} - \frac{37}{210}$ $(k+1)$

$=$ $\frac{1}{7}$ $[ k^7 + 7k^6+ 21k^5 + 35k^4 + 35 k^3 + 21 k^2 + 7k + 1] +$ $\frac{1}{5}$ $[ k^5 + 5k^4+10k^3+10k^2+5k+1] +$ $\frac{1}{3}$ $[ k^3 + 3k^2 + 3k + 1] +$ $\frac{1}{2}$ $[k^2 + 2k + 1] -$ $\frac{37}{210}$ $(k+1)$

$= \Big[$ $\frac{k^7}{7} + \frac{k^5}{5} + \frac{k^3}{3} + \frac{k^2}{2} - \frac{37k}{210}$ $\Big] +$ $\Big[ k^6 + 3k^5 + 5k^4 + 5k^3 + 3k^2 + k +\frac{1}{7} + k^4 + 2k^3 + 2k^2 + \frac{1}{5} + k^2 + k + \frac{1}{3}+ k + \frac{1}{2} - \frac{37}{210} \Big]$

$= \lambda + [ k^6 + 3k^5 + 5k^4 + 5k^3 + 3k^2 + k + k^4 + 2k^3 + 2k^2 + k^2 + k + k ] + \Big[ \frac{1}{7} + \frac{1}{5} + \frac{1}{3} + \frac{1}{2} - \frac{37}{210} \Big]$

$= \lambda + [k^6 + 3k^5 + 6k^4 + 7k^3 + 6k^2 + 3k] + 1$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$\frac{n^7}{7}$ $+$ $\frac{n^5}{5}$ $+$ $\frac{n^3}{3}$ $+$ $\frac{n^2}{2}$ $-$ $\frac{37}{210}$ $n$ is a positive integer got all $n \in N$

$\\$

Question 33: Prove that $\frac{n^{11}}{11}$ $+$ $\frac{n^5}{5}$ $+$ $\frac{n^3}{3}$ $+$ $\frac{62}{165}$ $n$ is a positive integer got all $n \in N$

Answer:

Let $P(n) :$ $\frac{n^{11}}{11}$ $+$ $\frac{n^5}{5}$ $+$ $\frac{n^3}{3}$ $+$ $\frac{62}{165}$ $n$ is a positive integer got all $n \in N$

For $n = 1$ ,

L.H.S $= \frac{1}{11}+\frac{1}{5}+\frac{1}{3}+\frac{62}{165} = \frac{15+33+55+62}{165}$ $= 1$

1 is an integer. Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$\frac{k^{11}}{11}$ $+$ $\frac{k^5}{5}$ $+$ $\frac{k^3}{3}$ $+$ $\frac{62}{165}$ $k$ is a positive integer got all $k \in N$

Now we have to show that $P(n)$ is true for $n = k+1$

$\frac{(k+1)^{11}}{11} + \frac{(k+1)^5}{5} + \frac{(k+1)^3}{3} + \frac{62}{165}$ $(k+1)$

$= \frac{1}{11} [ k^{11}+ 11k^{10}+55k^9 + 165k^8 +330k^7+ 462k^6+ 462k^5 + 330k^4 + 165k^3+ 55k^2+11k+1] + \frac{1}{5} [ k^5+5k^4+10k^3+10k^2+5k+1 ] + \frac{1}{3} [ k^3+3k^2+3k+1 ] + \frac{62}{165} [k+1]$

$= \Big[ \frac{k^{11}}{11} + \frac{k^{5}}{5} + \frac{k^{3}}{3} + \frac{62k}{165} \Big] + k^{10} + 5k^9 + 15k^8 + 30 k^7 + 42k^6 + 42k^5 + 30k^4 + 15k^3 + 5k^2 +1 + \frac{1}{11} + k^4 + 2k^3 + 2k^2 + k + \frac{1}{5} + k^2 + k + \frac{1}{3} + \frac{62}{165}$

$= \Big[ \frac{k^{11}}{11} + \frac{k^{5}}{5} + \frac{k^{3}}{3} + \frac{62k}{165} \Big] + k^{10} + 5k^9 + 15k^8 + 30k^7 + 42k^6 + 42k^5 + 31k^4 + 17k^3 + 8k^2 + 2k + 1$

Which is an integer. $\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$\frac{n^{11}}{11}$ $+$ $\frac{n^5}{5}$ $+$ $\frac{n^3}{3}$ $+$ $\frac{62}{165}$ $n$ is a positive integer got all $n \in N$

$\\$

Question 34: Prove that

$\frac{1}{2}$ $\tan \Big($ $\frac{x}{2}$ $\Big) +$ $\frac{1}{4}$ $\tan \Big($ $\frac{x}{4}$ $\Big) + \ldots +$ $\frac{1}{2^n}$ $\tan \Big($ $\frac{x}{2^n}$ $\Big) =$ $\frac{1}{2^n}$ $\cot \Big($ $\frac{x}{2^n}$ $\Big) - \cot x$

for all $n \in N$ and $0 < x <$ $\frac{\pi}{2}$

Answer:

Let $P(n) :$

for all $n \in N$ and $0 < x <$ $\frac{\pi}{2}$

For $n = 1$ , L.H.S $=$ $\frac{1}{2}$ $\tan$ $\frac{x}{2}$

R.H.S $=$ $\frac{1}{2}$ $\cot$ $\frac{x}{2}$ $-\cot x =$ $\frac{1}{2 \tan \frac{x}{2}}$ $-$ $\frac{1}{\tan x}$

$= \frac{1}{2 \tan \frac{x}{2}} - \frac{1}{\frac{2 \tan \frac{x}{2}}{1 - \tan^2 \frac{x}{2}}}$

$= \frac{1-1+\tan^2 \frac{x}{2}}{2 \tan \frac{x}{2}}$ $=$ $\frac{1}{2}$ $\tan$ $\frac{x}{2}$

Therefore LHS = RHS

Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$\frac{1}{2}$ $\tan \Big($ $\frac{x}{2}$ $\Big) +$ $\frac{1}{4}$ $\tan \Big($ $\frac{x}{4}$ $\Big) + \ldots +$ $\frac{1}{2^k}$ $\tan \Big($ $\frac{x}{2^k}$ $\Big) =$ $\frac{1}{2^k}$ $\cot \Big($ $\frac{x}{2^k}$ $\Big) - \cot x$ … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$\frac{1}{2} \tan \big( \frac{x}{2} \big) + \frac{1}{4} \tan \big( \frac{x}{4} \big) + \ldots + \frac{1}{2^k} \tan \big( \frac{x}{2^k} \big) + \frac{1}{2^{k+1}} \tan \big( \frac{x}{2^{k+1}} \big)$

$= \frac{1}{2^k} \cot \big( \frac{x}{2^k} \big) - \cot x + \frac{1}{2^{k+1}} \tan \big( \frac{x}{2^{k+1}} \big)$

$= \frac{1}{2^k} \Big[ \cot \big( \frac{x}{2^k} \big) + \frac{1}{2} \tan \big( \frac{x}{2^{k+1}} \big) \Big] - \cot x$

$= \frac{1}{2^k} \Big[ \frac{1}{\tan \big( \frac{x}{2^k} \big)} + \frac{1}{2} \tan \big( \frac{x}{2^{k+1}} \big) \Big] - \cot x$

$= \frac{1}{2^k} \Big[ \frac{1}{\tan \big( \frac{x}{2^{k+1}} + \frac{x}{2^{k+1}} \big)} + \frac{1}{2} \tan \big( \frac{x}{2^{k+1}} \big) \Big] - \cot x$

$= \frac{1}{2^k} \Big[ \frac{1 - \tan^2 \frac{x}{2^{k+1}}}{2 \tan \frac{x}{2^{k+1}}} + \frac{1}{2} \tan \big( \frac{x}{2^{k+1}} \big) \Big] - \cot x$

$= \frac{1}{2^{k+1} } \Big[ \frac{1 - \tan^2 \frac{x}{ 2^{k+1} } + \tan^2 \frac{x}{ 2^{k+1} }}{\tan^2 \frac{x}{ 2^{k+1} } } \Big] - \cot x$

$= \frac{1}{2^{k+1} } \cot^2 \frac{x}{ 2^{k+1} } - \cot x$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$\\$

Question 35: Prove that

$\Big( 1 -$ $\frac{1}{2^2}$ $\Big) \Big( 1 -$ $\frac{1}{3^2}$ $\Big) \Big( 1 -$ $\frac{1}{4^2}$ $\Big) \ldots \Big( 1 -$ $\frac{1}{n^2}$ $\Big) =$ $\frac{n+1}{2n}$ for all natural numbers, $n \geq 2$

Answer:

Let $P(n) :$ $\Big( 1 -$ $\frac{1}{2^2}$ $\Big) \Big( 1 -$ $\frac{1}{3^2}$ $\Big) \Big( 1 -$ $\frac{1}{4^2}$ $\Big) \ldots \Big( 1 -$ $\frac{1}{n^2}$ $\Big) =$ $\frac{n+1}{2n}$ for all natural numbers, $n \geq 2$

For $n = 2$ , L.H.S $= \big( 1 -$ $\frac{1}{2^2}$ $\big) =$ $\frac{3}{4}$ R.H.S $=$ $\frac{2+1}{2 \cdot 2}$ $=$ $\frac{3}{4}$ Therefore LHS = RHS

Hence $P(2)$ is true for $n = 2$

Let $P(n)$ is true for $n = k$

$\Big( 1 -$ $\frac{1}{2^2}$ $\Big) \Big( 1 -$ $\frac{1}{3^2}$ $\Big) \Big( 1 -$ $\frac{1}{4^2}$ $\Big) \ldots \Big( 1 -$ $\frac{1}{k^2}$ $\Big) =$ $\frac{k+1}{2k}$ for all natural numbers, $k \geq 2$ … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$\Big( 1 -$ $\frac{1}{2^2}$ $\Big) \Big( 1 -$ $\frac{1}{3^2}$ $\Big) \Big( 1 -$ $\frac{1}{4^2}$ $\Big) \ldots \Big( 1 -$ $\frac{1}{k^2}$ $\Big) \Big( 1 -$ $\frac{1}{(k+1)^2}$ $\Big)$

$= \Big($ $\frac{k+1}{2k}$ $\Big) \Big( 1 -$ $\frac{1}{(k+1)^2}$ $\Big)$

$=$ $\frac{k+1}{2k}$ $\Big($ $\frac{k^2+1 + 2k - 1}{(k+1)^2}$ $\Big)$

$=$ $\frac{k(k+2)}{2k ( k+1)}$

$=$ $\frac{k+2}{2(k+1)}$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$\Big( 1 -$ $\frac{1}{2^2}$ $\Big) \Big( 1 -$ $\frac{1}{3^2}$ $\Big) \Big( 1 -$ $\frac{1}{4^2}$ $\Big) \ldots \Big( 1 -$ $\frac{1}{n^2}$ $\Big) =$ $\frac{n+1}{2n}$ for all natural numbers, $n \geq 2$

$\\$

Question 36: Prove that $\frac{ (2n)!}{2^{2n}(n!)^2}$ $\leq$ $\frac{1}{\sqrt{3n+1}}$ for all $n \in N$

Answer:

Let $P(n) :$ $\frac{ (2n)!}{2^{2n}(n!)^2}$ $\leq$ $\frac{1}{\sqrt{3n+1}}$ for all $n \in N$

For $n = 1$ , $=$ $\frac{2!}{2^2 \cdot 1}$ $=$ $\frac{1}{2}$ $\leq$ $\frac{1}{\sqrt{3+1}}$

Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$\frac{ (2k)!}{2^{2k}(k!)^2}$ $\leq$ $\frac{1}{\sqrt{3k+1}}$ for all $n \in N$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$\frac{ [2(k+1)]!}{2^{2(k+1)}[(k+1)!]^2}$

$=$ $\frac{(2k+2)(2k+1)(2k)!}{2^{2k} \cdot 2^2 \cdot ( k+1) k! (k+1) k!}$

$=$ $\frac{(2k+1) (2k)!}{2^{2k} \cdot 2 k! (k+1) k!}$

$=$ $\frac{(2k)!}{2^{2k} (k!)^2}$ $\Big[$ $\frac{2k+1}{2 (k+1)}$ $\Big]$

$\leq$ $\frac{1}{\sqrt{3k+1}}$ $\Big[$ $\frac{2k+1}{2 (k+1)}$ $\Big]$

Since $2k+1 < 2k+2 \Rightarrow$ $\frac{2k+1}{2(k+1)}$ $<$ $\frac{2k+2}{2(k+1)}$ $= 1$

$\leq$ $\frac{1}{\sqrt{3k+1}}$

Since $3k+1 < 3k+4$

$\leq$ $\frac{1}{\sqrt{3k+4}}$

$\leq$ $\frac{1}{\sqrt{3(k+1)+1}}$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$\frac{ (2n)!}{2^{2n}(n!)^2}$ $\leq$ $\frac{1}{\sqrt{3n+1}}$ for all $n \in N$

$\\$

Question 37: $1 +$ $\frac{1}{4}$ $+$ $\frac{1}{9}$ $+$ $\frac{1}{16}$ $+ \ldots +$ $\frac{1}{n^2}$ $< 2 -$ $\frac{1}{n}$ for all $n \geq 2, n \in N$

Answer:

Let $P(n) :$ $1 +$ $\frac{1}{4}$ $+$ $\frac{1}{9}$ $+$ $\frac{1}{16}$ $+ \ldots +$ $\frac{1}{n^2}$ $< 2 -$ $\frac{1}{n}$ for all $n \geq 2, n \in N$

For $n = 2$ , $\frac{1}{2^2}$ $=$ $\frac{1}{4}$ $< 2 -$ $\frac{1}{2}$

Hence $P(2)$ is true for $n = 2$

Let $P(n)$ is true for $n = k$

$1 +$ $\frac{1}{4}$ $+$ $\frac{1}{9}$ $+$ $\frac{1}{16}$ $+ \ldots +$ $\frac{1}{k^2}$ $< 2 -$ $\frac{1}{k}$ for all $k \geq 2, n \in N$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$1 +$ $\frac{1}{4}$ $+$ $\frac{1}{9}$ $+$ $\frac{1}{16}$ $+ \ldots +$ $\frac{1}{k^2}$ $+$ $\frac{1}{(k+1)^2}$

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

$< 2 - \Big[$ $\frac{(k+1)^2 - k}{k(k+1)^2}$ $\Big]$

$< 2 - \Big[$ $\frac{(k^2 + 1 + 2k - k}{k(k+1)^2}$ $\Big]$

$< 2 - \Big[$ $\frac{(k^2 + k+1}{k(k+1)^2}$ $\Big]$

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

$< 2 -$ $\frac{1}{k+1}$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$1 +$ $\frac{1}{4}$ $+$ $\frac{1}{9}$ $+$ $\frac{1}{16}$ $+ \ldots +$ $\frac{1}{n^2}$ $< 2 -$ $\frac{1}{n}$ for all $n \geq 2, n \in N$

$\\$

Question 38: Prove that $x^{2n-1} + y^{2n-1}$ is divisible by $x+y$ for all $n \in N$

Answer:

Let $P(n) :$ $x^{2n-1} + y^{2n-1}$ is divisible by $x+y$ for all $n \in N$

For $n = 1 , x^{2-1}+y^{2-1} = x+y$ which is divisible by $x+y$

Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$x^{2k-1} + y^{2k-1}$ is divisible by $x+y$ for all $k \in N$

$\Rightarrow x^{2k-1} + y^{2k-1} = (x+y) \mu$ … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$x^{2(k+1) - 1} - y^{2(k+1) - 1}$

$= x^{2k+1} - y^{2k+1}$

$= x^{2k -1 +2} - y^{2k- 1+2}$

$= x^2 \cdot x^{2k-1} - y^2 \cdot y^{2k-1}$

$= x^2 [ (x+y) \mu + y^{2k-1} ] - y^2 \cdot y^{2k-1}$

$= x^2 (x+y) \mu + x^2 y^{2k-1} - y^2 \cdot y^{2k-1}$

$= x^2 (x+y) \mu + (x^2 - y^2) y^{2k-1}$

$= x^2 (x+y) \mu + (x+y)(x-y) y^{2k-1}$

$= (x+y) [ x^2 \mu + (x-y) y^{2k-1} ]$

$= (x+y) \lambda$ where $\lambda = [ x^2 \mu + (x-y) y^{2k-1} ]$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$x^{2n-1} + y^{2n-1}$ is divisible by $x+y$ for all $n \in N$

$\\$

Question 39: Prove that $\sin x + \sin 3x + \ldots + \sin (2n-1) x =$ $\frac{\sin^2 nx}{\sin x}$ for all $n \in N$

Answer:

Let $P(n) :$ $\sin x + \sin 3x + \ldots + \sin (2n-1) x =$ $\frac{\sin^2 nx}{\sin x}$ for all $n \in N$

For $n = 1 , LHS = \sin x RHS =$ $\frac{\sin^2 x}{\sin x}$ $= \sin x Therefore LHS = RHS Hence$ latex P(1) $ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$\sin x + \sin 3x + \ldots + \sin (2k-1) x =$ $\frac{\sin^2 kx}{\sin x}$ for all $k \in N$ … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$\sin x + \sin 3x + \ldots + \sin (2k-1) x + \sin [2(k+1) - 1]x$

$=$ $\frac{\sin^2 kx}{\sin x}$ $+ \sin [2(k+1) - 1]x$

$=$ $\frac{\sin^2 kx}{\sin x}$ $+ \sin (2k+1) x$

$=$ $\frac{\sin^2 kx + \sin x \cdot \sin (2k+1) x}{\sin x}$

$=$ $\frac{\sin^2 kx + \{ \cos [ (2k+1)x - x] - \cos [ (2k+1)x + x ] \} }{2\sin x}$

$=$ $\frac{\sin^2 kx + \cos 2kx - \cos [2x(k+1)] }{2\sin x}$

$=$ $\frac{1 - \cos 2kx + \cos 2kx - \cos 2x(k+1) }{2\sin x}$

$=$ $\frac{1- \cos 2x(k+1) }{2\sin x}$

$=$ $\frac{2 \sin^2 x (k+1) }{2\sin x}$

$=$ $\frac{ \sin^2 x (k+1) }{\sin x}$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$\sin x + \sin 3x + \ldots + \sin (2n-1) x =$ $\frac{\sin^2 nx}{\sin x}$ for all $n \in N$

$\\$

Question 40: Prove that $\cos \alpha + \cos ( \alpha + \beta) + \cos ( \alpha + 2\beta) + \ldots + \cos ( \alpha + (n-1) \beta) = \frac{\cos \big\{ \alpha + \big( \frac{n-1}{2} \big) \beta \big\} \sin \big( \frac{n \beta}{2} \big) }{\sin \big( \frac{\beta}{2} \big) }$

Answer:

Let $P(n) :$

$\cos \alpha + \cos ( \alpha + \beta) + \cos ( \alpha + 2\beta) + \ldots + \cos ( \alpha + (n-1) \beta) = \frac{\cos \big\{ \alpha + \big( \frac{n-1}{2} \big) \beta \big\} \sin \big( \frac{n \beta}{2} \big) }{\sin \big( \frac{\beta}{2} \big) }$

For $n = 1 , LHS = \cos ( \alpha + (1-1) \beta) = \cos \alpha$

RHS $= \frac{\cos \big\{ \alpha + \big( \frac{1-1}{2} \big) \beta \big\} \sin \big( \frac{\beta}{2} \big) }{\sin \big( \frac{\beta}{2} \big) }$ $= \cos \alpha$

Therefore LHS = RHS

Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$\cos \alpha + \cos ( \alpha + \beta) + \cos ( \alpha + 2\beta) + \ldots + \cos ( \alpha + (k-1) \beta) = \frac{\cos \big\{ \alpha + \big( \frac{k-1}{2} \big) \beta \big\} \sin \big( \frac{k \beta}{2} \big) }{\sin \big( \frac{\beta}{2} \big) }$ … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$\cos \alpha + \cos ( \alpha + \beta) + \cos ( \alpha + 2\beta) + \ldots + \cos ( \alpha + (k-1) \beta) + \cos ( \alpha + k \beta)$

$= \frac{\cos \big\{ \alpha + \big( \frac{k-1}{2} \big) \beta \big\} \sin \big( \frac{k \beta}{2} \big) }{\sin \big( \frac{\beta}{2} \big) }$ $+ \cos ( \alpha + k \beta)$

$= \frac{\sin \frac{\beta}{2} \cos ( \alpha + k \beta) + \cos \big\{ \alpha + \big( \frac{k-1}{2} \big) \beta \big\} \sin \big( \frac{k \beta}{2} \big)}{\sin \big( \frac{\beta}{2} \big)}$

$= \frac{\frac{1}{2} \Big[ \sin \{ \alpha + \big( \frac{2k+1}{2} \big) \beta \} - \sin \{ \alpha + \big( \frac{2k-1}{2} \big) \beta \} \Big] + \cos \{ \alpha + \big( \frac{k-1}{2} \big) \beta \} \sin \frac{k\beta}{2} }{\sin \big( \frac{\beta}{2} \big)}$

$= \frac{\frac{1}{2} \Big[ \sin \{ \alpha + \big( \frac{2k+1}{2} \big) \beta \} - \sin \{ \alpha + \big( \frac{2k-1}{2} \big) \beta \} \Big] + \frac{1}{2} \Big[ \sin \{ \alpha + \big( \frac{2k-1}{2} \big) \beta \} + \sin \big( - \alpha + \frac{\beta}{2} \big) \Big] }{\sin \big( \frac{\beta}{2} \big)}$

$= \frac{\frac{1}{2} \Big[ \sin \{ \alpha + \big( \frac{2k+1}{2} \big) \beta \} + \sin \big( - \alpha + \frac{\beta}{2} \big) \Big] }{\sin \big( \frac{\beta}{2} \big)}$

$= \frac{\frac{1}{2} \Big[ \sin \alpha \cos \big( \frac{2k+1}{2} \big) \beta + \cos \alpha \sin \big( \frac{2k+1}{2} \big) \beta + \sin \frac{\beta}{2} \cos \alpha - \cos \frac{\beta}{2} \sin \alpha \Big] }{\sin \big( \frac{\beta}{2} \big)}$

$= \frac{\frac{1}{2} \Big[ \sin \alpha \Big\{ \cos \big( \frac{2k+1}{2} \big) \beta - \cos \frac{\beta}{2} \Big\} + \cos \alpha \Big\{ \sin \big( \frac{2k+1}{2} \big) \beta + \sin \frac{\beta}{2} \Big\} \Big] }{\sin \big( \frac{\beta}{2} \big)}$

$= \frac{\frac{1}{2} \Big[ -2\sin \alpha \Big\{ \sin \big( \frac{k+1}{2} \big) \beta \cdot \sin \frac{k\beta}{2} \Big\} + 2\cos \alpha \Big\{ \sin \big( \frac{k+1}{2} \big) \beta \cdot \sin \frac{k\beta}{2} \Big\} \Big] }{\sin \big( \frac{\beta}{2} \big)}$

$= \frac{ \Big[ -2\sin \alpha \Big\{ \sin \big( \frac{k+1}{2} \big) \beta \cdot \sin \frac{k\beta}{2} \Big\} + \cos \alpha \Big\{ \sin \big( \frac{k+1}{2} \big) \beta \cdot \sin \frac{k\beta}{2} \Big\} \Big] }{\sin \big( \frac{\beta}{2} \big)}$

$= \frac{ \sin \big( \frac{k+1}{2} \big) \beta \Big[ \cos \alpha \cdot \cos \frac{k\beta}{2} - \sin \alpha \cdot \sin \frac{k\beta}{2} \Big] }{\sin \big( \frac{\beta}{2} \big)}$

$= \frac{ \sin \big( \frac{k+1}{2} \big) \beta \cos \big( \alpha + \frac{k\beta}{2} \big) }{\sin \big( \frac{\beta}{2} \big)}$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$\\$

Question 41: Prove that $\frac{1}{n+1}$ $+$ $\frac{1}{n+2}$ $+ \ldots +$ $\frac{1}{2n}$ $>$ $\frac{13}{24}$ , for all natural numbers $n > 1$

Answer:

Let $P(n) :$ $\frac{1}{n+1}$ $+$ $\frac{1}{n+2}$ $+ \ldots +$ $\frac{1}{2n}$ $>$ $\frac{13}{24}$ , for all natural numbers $n > 1$

For $n = 2$ , L.H.S $= \frac{1}{2+1} + \frac{1}{2 \times 2}= \frac{1}{3} + \frac{1}{4} = \frac{14}{24} > \frac{13}{24} =$ R.H.S

Hence $P(2)$ is true for $n = 2$

Let $P(n)$ is true for $n = k$

$\frac{1}{k+1}$ $+$ $\frac{1}{k+2}$ $+ \ldots +$ $\frac{1}{2k}$ $>$ $\frac{13}{24}$ , for all natural numbers $k > 1$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$\frac{1}{k+2} + \ldots + \frac{1}{2k} + \frac{1}{2k+1} + \frac{1}{2k+2}$

$= - \frac{1}{k+1} + \frac{1}{k+1}\frac{1}{k+2} + \ldots + \frac{1}{2k} + \frac{1}{2k+1} + \frac{1}{2k+2}$

$> \frac{13}{24} - \frac{1}{k+1} + \frac{1}{2k+1} + \frac{1}{2k+2}$

$> \frac{13}{24} - \frac{1}{2(k+1)} + \frac{1}{2k+1}$

$> \frac{13}{24} + \frac{2(k+1)-(2k+1)}{2(k+1)(2k+1)}$

$> \frac{13}{24} + \frac{2k+2-2k-1}{2(k+1)(2k+1)}$

$> \frac{13}{24} + \frac{1}{2(k+1)(2k+1)}$

$> \frac{13}{24}$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$\frac{1}{n+1}$ $+$ $\frac{1}{n+2}$ $+ \ldots +$ $\frac{1}{2n}$ $>$ $\frac{13}{24}$ , for all natural numbers $n > 1$

$\\$

Question 42: Give

$a_1 =$ $\frac{1}{2}$ $\Big( a_0 +$ $\frac{A}{a_0}$ $\Big), a_2 =$ $\frac{1}{2}$ $\Big( a_1 +$ $\frac{A}{a_1}$ $\Big)$ and $a_{n+1} =$ $\frac{1}{2}$ $\Big( a_n +$ $\frac{A}{a_n}$ $\Big)$

for $n \geq 2$ , where $a > 0 , A > 0$

Prove that $\frac{a_n - \sqrt{A}}{a_n + \sqrt{A}}$ $= \Big($ $\frac{a_1 - \sqrt{A}}{a_1+\sqrt{A}}$ $\Big)^{2^{n-1}}$

Answer:

Let $P(n) :$ $\frac{a_n - \sqrt{A}}{a_n + \sqrt{A}}$ $= \Big($ $\frac{a_1 - \sqrt{A}}{a_1+\sqrt{A}}$ $\Big)^{2^{n-1}}$

For $n = 1$ LHS $=$ $\frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}}$

RHS $= \Big($ $\frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}}$ $\Big)^{2-1} =$ $\frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}}$

Therefore LHS = RHS

Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$\frac{a_k - \sqrt{A}}{a_k + \sqrt{A}}$ $= \Big($ $\frac{a_1 - \sqrt{A}}{a_1+\sqrt{A}}$ $\Big)^{2^{k-1}}$ is true for $k \leq 2$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

LHS $=$ $\frac{a_{k+1} - \sqrt{A}}{a_{k+1} + \sqrt{A}}$

$=$ $\frac{ \frac{1}{2} \big( a_k + \frac{A}{a_k} \big) - \sqrt{A}}{\frac{1}{2} \big( a_k + \frac{A}{a_k} \big) + \sqrt{A}}$

$=$ $\frac{\frac{1}{2} \big( \frac{{a_k}^2+A - 2 a_k \sqrt{A}}{a_k} \big) }{\frac{1}{2} \big( \frac{{a_k}^2+A + 2 a_k \sqrt{A}}{a_k} \big)}$

$=$ $\frac{{a_k}^2+A - 2a_k \sqrt{A}}{{a_k}^2+A + 2a_k \sqrt{A}}$

$=$ $\frac{(a_k - \sqrt{A})^2}{(a_k + \sqrt{A})^2}$

$= \Big($ $\frac{a_k - \sqrt{A}}{a_k + \sqrt{A}}$ $\Big)^2$

$= \Big[ \Big($ $\frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}}$ $\Big)^{2^{k-1}} \Big]^2$

$= \Big($ $\frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}}$ $\Big)^{2^k}$

RHS $= \Big($ $\frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}}$ $\Big)^{2^{k+1-1}}$

$= \Big($ $\frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}}$ $\Big)^{2^k}$

Therefore LHS = RHS

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$\frac{a_n - \sqrt{A}}{a_n + \sqrt{A}}$ $= \Big($ $\frac{a_1 - \sqrt{A}}{a_1+\sqrt{A}}$ $\Big)^{2^{n-1}}$

$\\$

Question 43: Let $P(n)$ be the statement: $2^n \geq 3n$ . If $P(r)$ is true, show that $P(r+1)$ is true. Do you conclude that $P(n)$ is true for all $n \in N$ ?

Answer:

Let $P(n) :$ $2^n \geq 3n$ for all $n \in N$

For $n = 1$ , LHS $= 2^1 = 2$ RHS $= 3 \times 1 = 3$

Therefore LHS $<$ RHS

Hence $P(1)$ is NOT true for $n = 1$

Therefore $P(n)$ is NOT true for all $n \in N$

$\\$

Question 44: Show that by the principle of mathematical induction that the sum $S_n$ of the $n$ terms of the series $1^2+ 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + 2 \times 6^2 + 7^2 + \ldots$ is given by

$S_n = \Bigg\{$ $\begin{array}{ll} \frac{n(n+1)^2}{2}, \text{if} \ n \ \text{is even} \\ \frac{n^2(n+1)}{2} , \text{if} \ n \ \text{is odd} \end{array}$

Answer:

Let $P(n) :$

$S_n$ of the $n$ terms of the series $1^2+ 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + 2 \times 6^2 + 7^2 + \ldots$ is given by

$S_n = \Bigg\{$ $\begin{array}{ll} \frac{n(n+1)^2}{2}, \text{if} \ n \ \text{is even} \\ \frac{n^2(n+1)}{2} , \text{if} \ n \ \text{is odd} \end{array}$

For $n = 2$ ,

L.H.S $S_2= 1^2 + 2 \times 2^2 = 9$

R.H.S $S_2=$ $\frac{2(3)^2}{2}$ $= 9$

Therefore LHS = RHS

Hence $P(n)$ is true for $n = 2$

For $n = 3$ ,

L.H.S $S_3= 1^2 + 2 \times 2^2 + 3^2 = 18$

R.H.S $S_3=$ $\frac{3^2(4)}{2}$ $= 18$

Therefore LHS = RHS

Hence $P(n)$ is true for $n = 3$

Let $P(n)$ is true for $n = k$

$S_k$ of the $k$ terms of the series $1^2+ 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + 2 \times 6^2 + 7^2 + \ldots$ is given by

$S_k = \Bigg\{$ $\begin{array}{ll} \frac{k(k+1)^2}{2}, \text{if} \ k \ \text{is even} \\ \frac{k^2(k+1)}{2} , \text{if} \ k \ \text{is odd} \end{array}$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

If $k$ is even, $(k+1)$ is odd

$\therefore S_{k+1} = 1^2+ 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + 2 \times 6^2 + 7^2 + \ldots + 2k^2 + ( k+1)^2$

$S_{k+1} = S_k + ( k+1)^2$

$\Rightarrow S_{k+1} = \frac{k(k+1)^2}{2} + ( k+1)^2 = (k+1)^2 \big( \frac{k+2}{2} \big)$

If $k$ is odd, $(k+1)$ is even

$\therefore S_{k+1} = 1^2+ 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + 2 \times 6^2 + 7^2 + \ldots + 2k^2 + 2( k+1)^2$

$S_{k+1} = S_k + 2( k+1)^2$

$\Rightarrow S_{k+1} = \frac{k^2(k+1)}{2} + 2( k+1)^2 = (k+1) \big( \frac{(k+2)^2}{2} \big)$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$S_n$ of the $n$ terms of the series $1^2+ 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + 2 \times 6^2 + 7^2 + \ldots$ is given by

$S_n = \Bigg\{$ $\begin{array}{ll} \frac{n(n+1)^2}{2}, \text{if} \ n \ \text{is even} \\ \frac{n^2(n+1)}{2} , \text{if} \ n \ \text{is odd} \end{array}$

$\\$

Question 45: Prove that the number of subsets of a set containing $n$ distinct elements is $2^n$ for all $n \in N$

Answer:

Let P(n) : The number of subsets of a set containing $n$ distinct elements is $2^n$ for all $n \in N$

For $n = 1$ LHS $=$ number of subsets of a set containing only $1$ element $= 2 ( \phi$ and the set itself $)$ RHS $= 2^1= 2$ .

Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

The number of subsets of a set containing $k$ distinct elements is $2^k$ for all $k \in N$

Now we have to show that $P(n)$ is true for $n = k+1$

Let $A = \{1 ,2, 3, \ldots, k, k+1 \}$

Now $A = \{1 ,2, 3, \ldots, k \} \cup \{m+1 \}$

Using i) we can say that $\{1 ,2, 3, \ldots, k \}$ has $2^k$ subset and $\{ m+1 \}$ as $2^1$ subset.

$\Rightarrow A$ has $2^k + 2 = s^{k+1}$ subsets.

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction The number of subsets of a set containing $n$ distinct elements is $2^n$ for all $n \in N$

$\\$

Question 46: A sequence $a_1, a_2, a_3, \ldots$ is defined by letting $a_1 = 3$ and $a_k = 7 a_{k-1}$ for all natural numbers $k \geq 2$ . Show that $a_n = 3 \cdot 7^{n-1}$ for all $n \in N$ .

Answer:

Given a sequence $a_1, a_2, a_3, \ldots$ is defined by letting $a_1 = 3$ and $a_k = 7 a_{k-1}$ for all natural numbers $k \geq 2$

Let $P(n) :$ $a_n = 3 \cdot 7^{n-1}$ for all $n \in N$

For $n = 1$ , L.H.S $= a_1 = 3 \cdot 7^{1-1} = 3 \cdot 1 = 1$

Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$a_k = 3 \cdot 7^{k-1}$ for all $k \in N$ … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$x_{k+1} =$ $\frac{x_{k+1-1}}{k}$

$a_{k+1} = 7 a_k$

$= 7 \cdot 3 \cdot 7^{k-1}$

$= 3 \cdot 7^{k-1+1}$

$= 3 \cdot 7^{(k+1)-1}$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction $a_n = 3 \cdot 7^{n-1}$ for all $n \in N$

$\\$

Question 47: A sequence $x_1, x_2, x_3, \ldots$ is defined by letting $x_1 = 2$ and $x_k =$ $\frac{x_{k-1}}{n}$ for all natural numbers $k, k \geq 2$ . Show that $x_n =$ $\frac{2}{n!}$ for all $n \in N$ .

Answer:

Given a sequence $x_1, x_2, x_3, \ldots$ is defined by letting $x_1 = 2$ and $x_k =$ $\frac{x_{k-1}}{n}$ for all natural numbers $k, k \geq 2$ .

Let $P(n) :$ $x_n =$ $\frac{2}{n!}$ for all $n \in N$ .

For $n = 1$ , L.H.S $= x_1 =$ $\frac{2}{1!}$ $= 2$

Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$x_k =$ $\frac{2}{k!}$ for all $k \in N$ … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$x_{k+1} =$ $\frac{x_{k+1-1}}{k}$

$=$ $\frac{x_k}{k}$

$=$ $\frac{2}{k \times k!}$

$=$ $\frac{2}{(k+1)!}$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction $x_n =$ $\frac{2}{n!}$ for all $n \in N$

$\\$

Question 48: A sequence $x_0, x_1, x_2, x_3, \ldots$ is defined by letting $x_0 = 5$ and $x_k = 4+x_{k-1}$ for all natural numbers $k$ . Show that $x_n = 5+4n$ for all $n \in N$ using mathematical induction.

Answer:

Given A sequence $x_0, x_1, x_2, x_3, \ldots$ is defined by letting $x_0 = 5$ and $x_k = 4+x_{k-1}$ for all natural numbers $k$ .

Let $P(n) :$ $x_n = 5+4n$ for all $n \in N$

For $n = 0$ , L.H.S $= x_0 = 5 + 4 \times 0 = 5$

Hence $P(0)$ is true for $n = 0$

Let $P(n)$ is true for $n = k$

$x_k = 5+4k$ for all $n \in N$ … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$x_{k+1} = 4 + x_{k+1-1}$

$= 4 + x_{k}$

$= 4 + 5 + 4k$

$= 5 + 4 ( k+1)$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$x_n = 5+4n$ for all $n \in N$

$\\$

Question 49: Using the principle of mathematical induction prove that

$\sqrt{n} <$ $\frac{1}{\sqrt{1}}$ $+$ $\frac{1}{\sqrt{2}}$ $+$ $\frac{1}{\sqrt{3}}$ $+ \ldots +$ $\frac{1}{\sqrt{n}}$ for all natural numbers $n \geq 2$ .

Answer:

Let $P(n) :$ $\sqrt{n} <$ $\frac{1}{\sqrt{1}}$ $+$ $\frac{1}{\sqrt{2}}$ $+$ $\frac{1}{\sqrt{3}}$ $+ \ldots +$ $\frac{1}{\sqrt{n}}$ for all natural numbers $n \geq 2$

For $n = 2$ , L.H.S $= \sqrt{2} = 1.41$ R.H.S $= \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} = 1 + 0.707 = 1.707$ Therefore LHS < RHS

Hence $P(2)$ is true for $n = 2$

Let $P(n)$ is true for $n = k$

$\sqrt{k} <$ $\frac{1}{\sqrt{1}}$ $+$ $\frac{1}{\sqrt{2}}$ $+$ $\frac{1}{\sqrt{3}}$ $+ \ldots +$ $\frac{1}{\sqrt{k}}$ for all natural numbers $k \geq 2$ … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

LHS $= \sqrt{k+1}$

RHS $=$ $\frac{1}{\sqrt{1}}$ $+$ $\frac{1}{\sqrt{2}}$ $+$ $\frac{1}{\sqrt{3}}$ $+ \ldots +$ $\frac{1}{\sqrt{k}}$ $+$ $\frac{1}{\sqrt{k+1}}$

$> \sqrt{k} +$ $\frac{1}{\sqrt{k+1}}$

Since $\sqrt{k+1} > \sqrt{k}$

$\Rightarrow$ $\frac{\sqrt{k}}{\sqrt{k+1}}$ $< 1$

$\Rightarrow$ $\frac{1}{\sqrt{k+1}}$ $< \sqrt{k}$

$\Rightarrow$ $\frac{k+1}{\sqrt{k+1}}$ $-$ $\frac{1}{\sqrt{k+1}}$ $< \sqrt{k}$

$\Rightarrow \sqrt{k+1} -$ $\frac{1}{\sqrt{k+1}}$ $< \sqrt{k}$

$\Rightarrow \sqrt{k} +$ $\frac{1}{\sqrt{k+1}}$ $> \sqrt{k+1}$

Therefore LHS $<$ RHS

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$\sqrt{n} <$ $\frac{1}{\sqrt{1}}$ $+$ $\frac{1}{\sqrt{2}}$ $+$ $\frac{1}{\sqrt{3}}$ $+ \ldots +$ $\frac{1}{\sqrt{n}}$ for all natural numbers $n \geq 2$

$\\$

Question 50: The distributive law from algebra states that for all real numbers $c, a_1$ and $a_2$ we have $c(a_1+a_2) = ca_1 + ca_2$ . Use this law and mathematical induction to prove that, for all natural numbers, $n \geq 2$ , if $c, a_1, a_2, a_3, \ldots , a_n$ are any real numbers, then

$c(a_1+a_2+ \ldots + a_n) = ca_1 + ca_2 + \ldots + ca_n$

Answer:

For all real numbers $c, a_1$ , and $a_2, c( a_1 + a_2) = ca_1 + ca_2$

To prove that, for all natural numbers, $n \geq 2$ , if $c, a_1, a_2, a_3, \ldots , a_n$ are any real numbers, then $c(a_1+a_2+ \ldots + a_n) = ca_1 + ca_2 + \ldots + ca_n$

Let $P(n) :$ $c(a_1+a_2+ \ldots + a_n) = ca_1 + ca_2 + \ldots + ca_n$ for all natural numbers $n \geq 2$ , and $c, a_1, a_2, \ldots , a_n \in R$

For $n = 2$ , LHS $= c( a_1+a_2)$ , RHS $= ca_1 +ca_2 \Rightarrow$ LHS $=$ RHS

Hence $P(n)$ is true for $n=2$

Let $P(n)$ is true for $n = k$ therefore $c(a_1+a_2+ \ldots + a_k) = ca_1 + ca_2 + \ldots + ca_k$ for all natural numbers $n \geq 2$ , and $c, a_1, a_2, \ldots , a_k \in R$