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

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

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

$\displaystyle \text{Hence } P(1) \text{is true for} n = 1$

$\displaystyle \text{Let } P(n) \text{is true for} n = k$

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

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

$\displaystyle \text{Now we have to show that } P(n) \text{is true for} n = k+1$

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

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

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

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

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

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

We know that $\displaystyle 5^k - 5 \text{ is divisible by } 4$.

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

$\displaystyle \Rightarrow P(n) \text{is true for} n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

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

$\displaystyle \text{Hence } P(1) \text{is true for} n = 1$

$\displaystyle \text{Let } P(n) \text{is true for} n = k$

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

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

$\displaystyle \text{Now we have to show that } P(n) \text{is true for} n = k+1$

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

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

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{is true for} n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

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

$\displaystyle \text{Hence } P(1) \text{is true for} n = 1$

$\displaystyle \text{Let } P(n) \text{is true for} n = k$

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

$\displaystyle \text{Now we have to show that } P(n) \text{is true for} n = k+1$

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{is true for} n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

$\displaystyle \text{For } n = 1 , \text{L.H.S } = 1 - 7 + 3 = -3 \text{which is divisible by } 3$

$\displaystyle \text{Hence } P(1) \text{is true for} n = 1$

$\displaystyle \text{Let } P(n) \text{is true for} n = k$

$\displaystyle k^3 - 7k+3 \text{ is divisible by } 3 \text{ for all } k \in N$

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

$\displaystyle \text{Now we have to show that } P(n) \text{is true for} n = k+1$

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{is true for} n = k+1$

Hence by the principle of mathematical induction

$\displaystyle n^3 - 7n+3 \text{ is divisible by } 3 \text{ for all } n \in N$

$\displaystyle \\$

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

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

$\displaystyle \text{For } n = 1 , \text{L.H.S } = 1+ 2^1 = 3$ R.H.S $\displaystyle = 2^2 - 1 = 3 \text{ Therefore LHS = RHS }$

$\displaystyle \text{Hence } P(1) \text{is true for} n = 1$

$\displaystyle \text{Let } P(n) \text{is true for} n = k$

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

$\displaystyle \text{Now we have to show that } P(n) \text{is true for} n = k+1$

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{is true for} n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

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

$\displaystyle \text{Hence } P(1) \text{is true for} n = 1$

$\displaystyle \text{Let } P(n) \text{is true for} n = k$

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

$\displaystyle \text{Now we have to show that } P(n) \text{is true for} n = k+1$

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

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

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

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{is true for} n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

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

$\displaystyle \text{Hence } P(1) \text{is true for} n = 1$

$\displaystyle \text{Let } P(n) \text{is true for} n = k$

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

$\displaystyle \text{Now we have to show that } P(n) \text{is true for} n = k+1$

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{is true for} n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

$\displaystyle \text{For } n = 1$,

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

1 is an integer. $\displaystyle \text{Hence } P(1) \text{is true for} n = 1$

$\displaystyle \text{Let } P(n) \text{is true for} n = k$

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

$\displaystyle \text{Now we have to show that } P(n) \text{is true for} n = k+1$

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

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

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

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

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

Hence by the principle of mathematical induction

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

$\displaystyle \\$

Question 34: Prove that

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

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

$\displaystyle \text{Let } P(n) :$

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

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

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

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

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

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

Therefore LHS = RHS

$\displaystyle \text{Hence } P(1) \text{is true for} n = 1$

$\displaystyle \text{Let } P(n) \text{is true for} n = k$

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

$\displaystyle \text{Now we have to show that } P(n) \text{is true for} n = k+1$

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

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

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

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{is true for} n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

Question 35: Prove that

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

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

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

$\displaystyle \text{Hence } P(2) \text{is true for} n = 2$

$\displaystyle \text{Let } P(n) \text{is true for} n = k$

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

$\displaystyle \text{Now we have to show that } P(n) \text{is true for} n = k+1$

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{is true for} n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

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

$\displaystyle \text{Hence } P(1) \text{is true for} n = 1$

$\displaystyle \text{Let } P(n) \text{is true for} n = k$

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

$\displaystyle \text{Now we have to show that } P(n) \text{is true for} n = k+1$

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

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

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

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

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

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{is true for} n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

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

$\displaystyle \text{Hence } P(2) \text{is true for} n = 2$

$\displaystyle \text{Let } P(n) \text{is true for} n = k$

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

$\displaystyle \text{Now we have to show that } P(n) \text{is true for} n = k+1$

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

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

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{is true for} n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

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

$\displaystyle \text{Hence } P(1) \text{is true for} n = 1$

$\displaystyle \text{Let } P(n) \text{is true for} n = k$

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

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

$\displaystyle \text{Now we have to show that } P(n) \text{is true for} n = k+1$

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

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

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

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

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

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{is true for} n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

$\displaystyle \text{For } n = 1 , LHS = \sin x \ \ \ \ RHS = \frac{\sin^2 x}{\sin x} = \sin x$

Therefore LHS = RHS

$\displaystyle \text{Hence } P(1) \text{is true for} n = 1$

$\displaystyle \text{Let } P(n) \text{is true for} n = k$

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

$\displaystyle \text{Now we have to show that } P(n) \text{is true for} n = k+1$

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

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

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

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

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

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{is true for} n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

$\displaystyle \text{Let } P(n) :$

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

$\displaystyle \text{For } n = 1 , LHS = \cos ( \alpha + (1-1) \beta) = \cos \alpha$

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

Therefore LHS = RHS

$\displaystyle \text{Hence } P(1) \text{is true for} n = 1$

$\displaystyle \text{Let } P(n) \text{is true for} n = k$

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

$\displaystyle \text{Now we have to show that } P(n) \text{is true for} n = k+1$

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{is true for} n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

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

$\displaystyle \text{Hence } P(2) \text{is true for} n = 2$

$\displaystyle \text{Let } P(n) \text{is true for} n = k$

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

$\displaystyle \text{Now we have to show that } P(n) \text{is true for} n = k+1$

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

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

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

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

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

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

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

$\displaystyle > \frac{13}{24}$

$\displaystyle \Rightarrow P(n) \text{is true for} n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

Question 42: Give

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

$\displaystyle \text{For } n \geq 2 , \text{ where } a > 0 , A > 0$

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

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

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

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

Therefore LHS = RHS

$\displaystyle \text{Hence } P(1) \text{is true for} n = 1$

$\displaystyle \text{Let } P(n) \text{is true for} n = k$

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

$\displaystyle \text{Now we have to show that } P(n) \text{is true for} n = k+1$

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

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

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

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

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

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

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

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

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

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

Therefore LHS = RHS

$\displaystyle \Rightarrow P(n) \text{is true for} n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

$\displaystyle \text{For } n = 1$, LHS $\displaystyle = 2^1 = 2$ RHS $\displaystyle = 3 \times 1 = 3$

Therefore LHS $\displaystyle <$ RHS

$\displaystyle \text{Hence } P(1)$ is NOT true $\displaystyle \text{For } n = 1$

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

$\displaystyle \\$

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

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

$\displaystyle \text{Let } P(n) :$

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

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

$\displaystyle \text{For } n = 2$,

$\displaystyle \text{L.H.S } S_2= 1^2 + 2 \times 2^2 = 9$

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

Therefore LHS = RHS

$\displaystyle \text{Hence } P(n) \text{is true for} n = 2$

$\displaystyle \text{For } n = 3$,

$\displaystyle \text{L.H.S } S_3= 1^2 + 2 \times 2^2 + 3^2 = 18$

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

Therefore LHS = RHS

$\displaystyle \text{Hence } P(n) \text{is true for} n = 3$

$\displaystyle \text{Let } P(n) \text{is true for} n = k$

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

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

$\displaystyle \text{Now we have to show that } P(n) \text{is true for} n = k+1$

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

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

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

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{is true for} n = k+1$

Hence by the principle of mathematical induction

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

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

$\displaystyle \\$

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

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

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

$\displaystyle \text{Hence } P(1) \text{is true for} n = 1$

$\displaystyle \text{Let } P(n) \text{is true for} n = k$

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

$\displaystyle \text{Now we have to show that } P(n) \text{is true for} n = k+1$

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{is true for} n = k+1$

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

$\displaystyle \\$

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

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

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

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

$\displaystyle \text{Hence } P(1) \text{is true for} n = 1$

$\displaystyle \text{Let } P(n) \text{is true for} n = k$

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

$\displaystyle \text{Now we have to show that } P(n) \text{is true for} n = k+1$

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

$\displaystyle a_{k+1} = 7 a_k$

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

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

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

$\displaystyle \Rightarrow P(n) \text{is true for} n = k+1$

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

$\displaystyle \\$

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

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

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

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

$\displaystyle \text{Hence } P(1) \text{is true for} n = 1$

$\displaystyle \text{Let } P(n) \text{is true for} n = k$

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

$\displaystyle \text{Now we have to show that } P(n) \text{is true for} n = k+1$

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{is true for} n = k+1$

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

$\displaystyle \\$

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

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

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

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

$\displaystyle \text{Hence } P(0) \text{is true for} n = 0$

$\displaystyle \text{Let } P(n) \text{is true for} n = k$

$\displaystyle x_k = 5+4k \text{ for all } n \in N$ … … i)

$\displaystyle \text{Now we have to show that } P(n) \text{is true for} n = k+1$

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

$\displaystyle = 4 + x_{k}$

$\displaystyle = 4 + 5 + 4k$

$\displaystyle = 5 + 4 ( k+1)$

$\displaystyle \Rightarrow P(n) \text{is true for} n = k+1$

Hence by the principle of mathematical induction

$\displaystyle x_n = 5+4n \text{ for all } n \in N$

$\displaystyle \\$

Question 49: Using the principle of mathematical induction prove that

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

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

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

$\displaystyle \text{Hence } P(2) \text{is true for} n = 2$

$\displaystyle \text{Let } P(n) \text{is true for} n = k$

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

$\displaystyle \text{Now we have to show that } P(n) \text{is true for} n = k+1$

LHS $\displaystyle = \sqrt{k+1}$

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

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

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

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

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

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

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

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

Therefore LHS $\displaystyle <$ RHS

$\displaystyle \Rightarrow P(n) \text{is true for} n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

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

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

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

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

$\displaystyle \text{Hence } P(n) \text{is true for} n=2$

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

$\displaystyle \text{For } n = k+1$

$\displaystyle c[a_1+a_2+ \ldots + a_k + a_{k+1}]$

$\displaystyle = c (a_1+a_2+ \ldots + a_k) + ca_{k+1}$

$\displaystyle = ca_1 + ca_2 + \ldots + ca_k + ca_{k+1}$

Therefore $\displaystyle P(n) \text{is true for} n = k+1$

Hence by the principle of mathematical induction

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