\displaystyle \textbf{Question 26: }\text{Prove by mathematical induction that }2\times7^n+3\times5^n-5\text{ is divisible}
\displaystyle \text{by }24\text{ for all }n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):2\times7^n+3\times5^n-5\text{ is divisible by }24.
\displaystyle \text{For }n=1,
\displaystyle 2\times7^1+3\times5^1-5=14+15-5=24
\displaystyle \text{Since }24\text{ is divisible by }24,\ P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore 2\times7^k+3\times5^k-5\text{ is divisible by }24.
\displaystyle \therefore 2\times7^k+3\times5^k-5=24\lambda,\text{ where }\lambda\in Z
\displaystyle \therefore 2\times7^k=24\lambda-3\times5^k+5\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle 2\times7^{k+1}+3\times5^{k+1}-5
\displaystyle =7(2\times7^k)+5(3\times5^k)-5
\displaystyle \text{Using equation (i),}
\displaystyle =7(24\lambda-3\times5^k+5)+5(3\times5^k)-5
\displaystyle =7\times24\lambda-21\times5^k+35+15\times5^k-5
\displaystyle =7\times24\lambda-6\times5^k+30
\displaystyle =7\times24\lambda-6(5^k-5)
\displaystyle \text{Since }5\equiv1\pmod 4,\text{ we have }5^k\equiv1\pmod 4.
\displaystyle \therefore 5^k-5\text{ is divisible by }4.
\displaystyle \therefore 5^k-5=4\mu,\text{ where }\mu\in Z
\displaystyle \therefore 2\times7^{k+1}+3\times5^{k+1}-5
\displaystyle =7\times24\lambda-6(4\mu)
\displaystyle =24(7\lambda-\mu)
\displaystyle \therefore 2\times7^{k+1}+3\times5^{k+1}-5\text{ is divisible by }24.
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle 2\times7^n+3\times5^n-5\text{ is divisible by }24\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{Prove by mathematical induction that }11^{n+2}+12^{2n+1}\text{ is divisible}
\displaystyle \text{by }133\text{ for all }n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):11^{n+2}+12^{2n+1}\text{ is divisible by }133.
\displaystyle \text{For }n=1,
\displaystyle 11^{1+2}+12^{2(1)+1}=11^3+12^3
\displaystyle =1331+1728=3059
\displaystyle =133\times23
\displaystyle \text{Therefore, }11^3+12^3\text{ is divisible by }133.
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore 11^{k+2}+12^{2k+1}\text{ is divisible by }133.
\displaystyle \therefore 11^{k+2}+12^{2k+1}=133\lambda,\text{ where }\lambda\in Z
\displaystyle \therefore 11^{k+2}=133\lambda-12^{2k+1}\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle 11^{(k+1)+2}+12^{2(k+1)+1}
\displaystyle =11^{k+3}+12^{2k+3}
\displaystyle =11\times11^{k+2}+144\times12^{2k+1}
\displaystyle \text{Using equation (i),}
\displaystyle =11(133\lambda-12^{2k+1})+144\times12^{2k+1}
\displaystyle =11\times133\lambda-11\times12^{2k+1}+144\times12^{2k+1}
\displaystyle =11\times133\lambda+133\times12^{2k+1}
\displaystyle =133(11\lambda+12^{2k+1})
\displaystyle \text{Let }\mu=11\lambda+12^{2k+1}.
\displaystyle \therefore 11^{(k+1)+2}+12^{2(k+1)+1}=133\mu
\displaystyle \therefore 11^{(k+1)+2}+12^{2(k+1)+1}\text{ is divisible by }133.
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle 11^{n+2}+12^{2n+1}\text{ is divisible by }133\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{Prove by mathematical induction that }1\times1!+2\times2!+3\times3!+\ldots
\displaystyle +n\times n!=(n+1)!-1\text{ for all }n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):1\times1!+2\times2!+3\times3!+\ldots+n\times n!=(n+1)!-1
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=1\times1!=1
\displaystyle \text{RHS}=(1+1)!-1=2!-1=1
\displaystyle \therefore \text{LHS}=\text{RHS}
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore 1\times1!+2\times2!+3\times3!+\ldots+k\times k!
\displaystyle =(k+1)!-1\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle \text{LHS of }P(k+1)=1\times1!+2\times2!+\ldots+k\times k!
\displaystyle +(k+1)\times(k+1)!
\displaystyle \text{Using equation (i),}
\displaystyle =(k+1)!-1+(k+1)\times(k+1)!
\displaystyle =(k+1)![1+(k+1)]-1
\displaystyle =(k+1)!(k+2)-1
\displaystyle =(k+2)!-1
\displaystyle =[(k+1)+1]!-1
\displaystyle =\text{RHS of }P(k+1)
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle 1\times1!+2\times2!+3\times3!+\ldots+n\times n!=(n+1)!-1
\displaystyle \text{for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{Prove by mathematical induction that }n^3-7n+3\text{ is divisible by }3
\displaystyle \text{for all }n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):n^3-7n+3\text{ is divisible by }3.
\displaystyle \text{For }n=1,
\displaystyle 1^3-7(1)+3=1-7+3=-3
\displaystyle \text{Since }-3\text{ is divisible by }3,\ P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore k^3-7k+3\text{ is divisible by }3.
\displaystyle \therefore k^3-7k+3=3\lambda,\text{ where }\lambda\in Z
\displaystyle \qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle (k+1)^3-7(k+1)+3
\displaystyle =k^3+3k^2+3k+1-7k-7+3
\displaystyle =(k^3-7k+3)+(3k^2+3k-6)
\displaystyle \text{Using equation (i),}
\displaystyle =3\lambda+3(k^2+k-2)
\displaystyle =3[\lambda+k^2+k-2]
\displaystyle \text{Let }\mu=\lambda+k^2+k-2.
\displaystyle \therefore (k+1)^3-7(k+1)+3=3\mu
\displaystyle \therefore (k+1)^3-7(k+1)+3\text{ is divisible by }3.
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle n^3-7n+3\text{ is divisible by }3\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{Prove by mathematical induction that }1+2+2^2+\ldots+2^n
\displaystyle =2^{n+1}-1\text{ for all }n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):1+2+2^2+\ldots+2^n=2^{n+1}-1
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=1+2^1=3
\displaystyle \text{RHS}=2^{1+1}-1=2^2-1=3
\displaystyle \therefore \text{LHS}=\text{RHS}
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore 1+2+2^2+\ldots+2^k=2^{k+1}-1\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle \text{LHS of }P(k+1)=1+2+2^2+\ldots+2^k+2^{k+1}
\displaystyle \text{Using equation (i),}
\displaystyle =2^{k+1}-1+2^{k+1}
\displaystyle =2\times2^{k+1}-1
\displaystyle =2^{k+2}-1
\displaystyle =2^{(k+1)+1}-1
\displaystyle =\text{RHS of }P(k+1)
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{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 \\

\displaystyle \textbf{Question 31: }\text{Prove by mathematical induction that }7+77+777+\ldots
\displaystyle +\underbrace{777\ldots7}_{n\text{ digits}}=\frac{7}{81}(10^{n+1}-9n-10)\text{ for all }n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):7+77+777+\ldots+\underbrace{777\ldots7}_{n\text{ digits}}
\displaystyle =\frac{7}{81}(10^{n+1}-9n-10)
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=7
\displaystyle \text{RHS}=\frac{7}{81}(10^{1+1}-9\times1-10)
\displaystyle =\frac{7}{81}(100-9-10)
\displaystyle =\frac{7}{81}\times81=7
\displaystyle \therefore \text{LHS}=\text{RHS}
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore 7+77+777+\ldots+\underbrace{777\ldots7}_{k\text{ digits}}
\displaystyle =\frac{7}{81}(10^{k+1}-9k-10)\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle \text{LHS of }P(k+1)=7+77+\ldots+\underbrace{777\ldots7}_{k\text{ digits}}
\displaystyle +\underbrace{777\ldots7}_{(k+1)\text{ digits}}
\displaystyle \text{Using equation (i),}
\displaystyle =\frac{7}{81}(10^{k+1}-9k-10)+7\underbrace{111\ldots1}_{(k+1)\text{ 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\times10^{k+1}-9)
\displaystyle =\frac{7}{81}[10^{k+1}-9k-10+9\times10^{k+1}-9]
\displaystyle =\frac{7}{81}[10\times10^{k+1}-9k-19]
\displaystyle =\frac{7}{81}[10^{k+2}-9(k+1)-10]
\displaystyle =\frac{7}{81}[10^{(k+1)+1}-9(k+1)-10]
\displaystyle =\text{RHS of }P(k+1)
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle 7+77+777+\ldots+\underbrace{777\ldots7}_{n\text{ digits}}
\displaystyle =\frac{7}{81}(10^{n+1}-9n-10)\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{Prove by mathematical induction that }\frac{n^7}{7}+\frac{n^5}{5}+\frac{n^3}{3}+\frac{n^2}{2}
\displaystyle -\frac{37n}{210}\text{ is a positive integer for all }n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):\frac{n^7}{7}+\frac{n^5}{5}+\frac{n^3}{3}+\frac{n^2}{2}-\frac{37n}{210}
\displaystyle \text{is a positive integer.}
\displaystyle \text{For }n=1,
\displaystyle \frac{1^7}{7}+\frac{1^5}{5}+\frac{1^3}{3}+\frac{1^2}{2}-\frac{37}{210}
\displaystyle =\frac{30+42+70+105-37}{210}
\displaystyle =\frac{210}{210}=1
\displaystyle \text{Since }1\text{ is a positive integer, }P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore \frac{k^7}{7}+\frac{k^5}{5}+\frac{k^3}{3}+\frac{k^2}{2}-\frac{37k}{210}=\lambda,
\displaystyle \text{where }\lambda\text{ is a positive integer.}\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle \frac{(k+1)^7}{7}+\frac{(k+1)^5}{5}+\frac{(k+1)^3}{3}+\frac{(k+1)^2}{2}
\displaystyle -\frac{37(k+1)}{210}
\displaystyle =\frac{1}{7}(k^7+7k^6+21k^5+35k^4+35k^3+21k^2+7k+1)
\displaystyle +\frac{1}{5}(k^5+5k^4+10k^3+10k^2+5k+1)
\displaystyle +\frac{1}{3}(k^3+3k^2+3k+1)+\frac{1}{2}(k^2+2k+1)
\displaystyle -\frac{37k}{210}-\frac{37}{210}
\displaystyle =\left(\frac{k^7}{7}+\frac{k^5}{5}+\frac{k^3}{3}+\frac{k^2}{2}-\frac{37k}{210}\right)
\displaystyle +k^6+3k^5+5k^4+5k^3+3k^2+k+\frac{1}{7}
\displaystyle +k^4+2k^3+2k^2+k+\frac{1}{5}
\displaystyle +k^2+k+\frac{1}{3}+k+\frac{1}{2}-\frac{37}{210}
\displaystyle \text{Using equation (i),}
\displaystyle =\lambda+k^6+3k^5+6k^4+7k^3+6k^2+4k
\displaystyle +\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{3}+\frac{1}{2}-\frac{37}{210}\right)
\displaystyle =\lambda+k^6+3k^5+6k^4+7k^3+6k^2+4k+1
\displaystyle \text{Since }\lambda\text{ is a positive integer and }k\in N,
\displaystyle \lambda+k^6+3k^5+6k^4+7k^3+6k^2+4k+1
\displaystyle \text{is also a positive integer.}
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{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{37n}{210}
\displaystyle \text{is a positive integer for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{Prove that }\frac{n^{11}}{11}+\frac{n^5}{5}+\frac{n^3}{3}+\frac{62n}{165}
\displaystyle \text{is a positive integer for all }n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n)\text{ be the statement that }\frac{n^{11}}{11}+\frac{n^5}{5}+\frac{n^3}{3}+\frac{62n}{165}
\displaystyle \text{is a positive integer.}
\displaystyle \text{For }n=1,
\displaystyle \frac{1^{11}}{11}+\frac{1^5}{5}+\frac{1^3}{3}+\frac{62}{165}
\displaystyle =\frac{15+33+55+62}{165}=\frac{165}{165}=1.
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \text{Thus, }\frac{k^{11}}{11}+\frac{k^5}{5}+\frac{k^3}{3}+\frac{62k}{165}\text{ is a positive integer.}
\displaystyle \text{Now, consider the expression for }n=k+1.
\displaystyle \frac{(k+1)^{11}}{11}+\frac{(k+1)^5}{5}+\frac{(k+1)^3}{3}+\frac{62(k+1)}{165}
\displaystyle =\frac{1}{11}\left(k^{11}+11k^{10}+55k^9+165k^8+330k^7+462k^6\right.
\displaystyle \left.+462k^5+330k^4+165k^3+55k^2+11k+1\right)
\displaystyle \quad+\frac{1}{5}\left(k^5+5k^4+10k^3+10k^2+5k+1\right)
\displaystyle \quad+\frac{1}{3}\left(k^3+3k^2+3k+1\right)+\frac{62(k+1)}{165}
\displaystyle =\left(\frac{k^{11}}{11}+\frac{k^5}{5}+\frac{k^3}{3}+\frac{62k}{165}\right)
\displaystyle \quad+k^{10}+5k^9+15k^8+30k^7+42k^6+42k^5
\displaystyle \quad+31k^4+17k^3+8k^2+3k
\displaystyle \quad+\left(\frac{1}{11}+\frac{1}{5}+\frac{1}{3}+\frac{62}{165}\right)
\displaystyle =\left(\frac{k^{11}}{11}+\frac{k^5}{5}+\frac{k^3}{3}+\frac{62k}{165}\right)
\displaystyle \quad+k^{10}+5k^9+15k^8+30k^7+42k^6+42k^5
\displaystyle \quad+31k^4+17k^3+8k^2+3k+1.
\displaystyle \text{By the induction hypothesis, the first bracket is a positive integer.}
\displaystyle \text{Also, }k^{10}+5k^9+15k^8+30k^7+42k^6+42k^5
\displaystyle \quad+31k^4+17k^3+8k^2+3k+1\text{ is a positive integer.}
\displaystyle \therefore \frac{(k+1)^{11}}{11}+\frac{(k+1)^5}{5}+\frac{(k+1)^3}{3}+\frac{62(k+1)}{165}
\displaystyle \text{is a positive integer. Hence, }P(k+1)\text{ is true.}
\displaystyle \text{Therefore, by the principle of mathematical induction,}
\displaystyle \frac{n^{11}}{11}+\frac{n^5}{5}+\frac{n^3}{3}+\frac{62n}{165}
\displaystyle \text{is a positive integer for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{Prove that}
\displaystyle \frac{1}{2}\tan\left(\frac{x}{2}\right)+\frac{1}{4}\tan\left(\frac{x}{4}\right)+\cdots
\displaystyle +\frac{1}{2^n}\tan\left(\frac{x}{2^n}\right)=\frac{1}{2^n}\cot\left(\frac{x}{2^n}\right)-\cot x,
\displaystyle \text{for all }n\in N\text{ and }0<x<\frac{\pi}{2}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n)\text{ be the statement}
\displaystyle \frac{1}{2}\tan\left(\frac{x}{2}\right)+\frac{1}{4}\tan\left(\frac{x}{4}\right)+\cdots
\displaystyle +\frac{1}{2^n}\tan\left(\frac{x}{2^n}\right)=\frac{1}{2^n}\cot\left(\frac{x}{2^n}\right)-\cot x.
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=\frac{1}{2}\tan\frac{x}{2}.
\displaystyle \text{RHS}=\frac{1}{2}\cot\frac{x}{2}-\cot x
\displaystyle =\frac{1}{2\tan\frac{x}{2}}-\frac{1}{\tan x}.
\displaystyle \text{Using }\tan x=\frac{2\tan\frac{x}{2}}{1-\tan^2\frac{x}{2}},
\displaystyle \text{RHS}=\frac{1}{2\tan\frac{x}{2}}-\frac{1-\tan^2\frac{x}{2}}{2\tan\frac{x}{2}}
\displaystyle =\frac{\tan^2\frac{x}{2}}{2\tan\frac{x}{2}}
\displaystyle =\frac{1}{2}\tan\frac{x}{2}=\text{LHS}.
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore \frac{1}{2}\tan\left(\frac{x}{2}\right)+\frac{1}{4}\tan\left(\frac{x}{4}\right)+\cdots
\displaystyle +\frac{1}{2^k}\tan\left(\frac{x}{2^k}\right)=\frac{1}{2^k}\cot\left(\frac{x}{2^k}\right)-\cot x.\qquad\text{...(i)}
\displaystyle \text{Now, consider the expression for }n=k+1.
\displaystyle \frac{1}{2}\tan\left(\frac{x}{2}\right)+\frac{1}{4}\tan\left(\frac{x}{4}\right)+\cdots
\displaystyle +\frac{1}{2^k}\tan\left(\frac{x}{2^k}\right)+\frac{1}{2^{k+1}}\tan\left(\frac{x}{2^{k+1}}\right)
\displaystyle =\frac{1}{2^k}\cot\left(\frac{x}{2^k}\right)-\cot x
\displaystyle \quad+\frac{1}{2^{k+1}}\tan\left(\frac{x}{2^{k+1}}\right)\qquad\text{[Using (i)]}
\displaystyle =\frac{1}{2^k}\left[\cot\left(\frac{x}{2^k}\right)+\frac{1}{2}\tan\left(\frac{x}{2^{k+1}}\right)\right]-\cot x.
\displaystyle \text{Let }\theta=\frac{x}{2^{k+1}}.\text{ Then }\frac{x}{2^k}=2\theta.
\displaystyle \cot 2\theta+\frac{1}{2}\tan\theta
\displaystyle =\frac{1-\tan^2\theta}{2\tan\theta}+\frac{1}{2}\tan\theta
\displaystyle =\frac{1-\tan^2\theta+\tan^2\theta}{2\tan\theta}
\displaystyle =\frac{1}{2\tan\theta}=\frac{1}{2}\cot\theta.
\displaystyle \therefore \cot\left(\frac{x}{2^k}\right)+\frac{1}{2}\tan\left(\frac{x}{2^{k+1}}\right)
\displaystyle =\frac{1}{2}\cot\left(\frac{x}{2^{k+1}}\right).
\displaystyle \therefore \frac{1}{2}\tan\left(\frac{x}{2}\right)+\frac{1}{4}\tan\left(\frac{x}{4}\right)+\cdots
\displaystyle +\frac{1}{2^{k+1}}\tan\left(\frac{x}{2^{k+1}}\right)
\displaystyle =\frac{1}{2^{k+1}}\cot\left(\frac{x}{2^{k+1}}\right)-\cot x.
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle \frac{1}{2}\tan\left(\frac{x}{2}\right)+\frac{1}{4}\tan\left(\frac{x}{4}\right)+\cdots
\displaystyle +\frac{1}{2^n}\tan\left(\frac{x}{2^n}\right)=\frac{1}{2^n}\cot\left(\frac{x}{2^n}\right)-\cot x
\displaystyle \text{for all }n\in N\text{ and }0<x<\frac{\pi}{2}.
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{Prove that}
\displaystyle \left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)\cdots
\displaystyle \left(1-\frac{1}{n^2}\right)=\frac{n+1}{2n}\text{ for all natural numbers }n\geq2.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n)\text{ be the statement}
\displaystyle \left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)\cdots
\displaystyle \left(1-\frac{1}{n^2}\right)=\frac{n+1}{2n},\qquad n\geq2.
\displaystyle \text{For }n=2,
\displaystyle \text{LHS}=1-\frac{1}{2^2}=1-\frac{1}{4}=\frac{3}{4}.
\displaystyle \text{RHS}=\frac{2+1}{2\times2}=\frac{3}{4}.
\displaystyle \therefore \text{LHS}=\text{RHS}.
\displaystyle \therefore P(2)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some natural number }k\geq2.
\displaystyle \therefore \left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)\cdots
\displaystyle \left(1-\frac{1}{k^2}\right)=\frac{k+1}{2k}.\qquad\text{...(i)}
\displaystyle \text{Now, consider the expression for }n=k+1.
\displaystyle \left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots
\displaystyle \left(1-\frac{1}{k^2}\right)\left(1-\frac{1}{(k+1)^2}\right)
\displaystyle =\frac{k+1}{2k}\left(1-\frac{1}{(k+1)^2}\right)\qquad\text{[Using (i)]}
\displaystyle =\frac{k+1}{2k}\left(\frac{(k+1)^2-1}{(k+1)^2}\right)
\displaystyle =\frac{k+1}{2k}\left(\frac{k^2+2k}{(k+1)^2}\right)
\displaystyle =\frac{k+1}{2k}\left(\frac{k(k+2)}{(k+1)^2}\right)
\displaystyle =\frac{k+2}{2(k+1)}.
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle \left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)\cdots
\displaystyle \left(1-\frac{1}{n^2}\right)=\frac{n+1}{2n}
\displaystyle \text{for all natural numbers }n\geq2.
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{Prove that }\frac{(2n)!}{2^{2n}(n!)^2}\leq\frac{1}{\sqrt{3n+1}}
\displaystyle \text{for all }n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n)\text{ be the statement}
\displaystyle \frac{(2n)!}{2^{2n}(n!)^2}\leq\frac{1}{\sqrt{3n+1}}.
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=\frac{2!}{2^2(1!)^2}=\frac{2}{4}=\frac{1}{2}.
\displaystyle \text{RHS}=\frac{1}{\sqrt{3(1)+1}}=\frac{1}{\sqrt{4}}=\frac{1}{2}.
\displaystyle \therefore \text{LHS}=\text{RHS}.
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore \frac{(2k)!}{2^{2k}(k!)^2}\leq\frac{1}{\sqrt{3k+1}}.\qquad\text{...(i)}
\displaystyle \text{Now, consider the expression 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+2}(k+1)^2(k!)^2}
\displaystyle =\frac{2(k+1)(2k+1)(2k)!}{4\cdot2^{2k}(k+1)^2(k!)^2}
\displaystyle =\frac{(2k)!}{2^{2k}(k!)^2}\cdot\frac{2k+1}{2(k+1)}
\displaystyle \leq\frac{1}{\sqrt{3k+1}}\cdot\frac{2k+1}{2(k+1)}\qquad\text{[Using (i)]}
\displaystyle =\frac{2k+1}{2(k+1)\sqrt{3k+1}}.
\displaystyle \text{We now prove that}
\displaystyle \frac{2k+1}{2(k+1)\sqrt{3k+1}}\leq\frac{1}{\sqrt{3k+4}}.
\displaystyle \text{Since both sides are positive, squaring both sides gives}
\displaystyle \frac{(2k+1)^2}{4(k+1)^2(3k+1)}\leq\frac{1}{3k+4}.
\displaystyle \text{This is equivalent to}
\displaystyle (2k+1)^2(3k+4)\leq4(k+1)^2(3k+1).
\displaystyle \text{Now,}
\displaystyle 4(k+1)^2(3k+1)-(2k+1)^2(3k+4)=k.
\displaystyle \text{Since }k\in N,\text{ we have }k>0.
\displaystyle \therefore (2k+1)^2(3k+4)<4(k+1)^2(3k+1).
\displaystyle \therefore \frac{2k+1}{2(k+1)\sqrt{3k+1}}\leq\frac{1}{\sqrt{3k+4}}.
\displaystyle \therefore \frac{[2(k+1)]!}{2^{2(k+1)}[(k+1)!]^2}
\displaystyle \leq\frac{1}{\sqrt{3k+4}}=\frac{1}{\sqrt{3(k+1)+1}}.
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle \frac{(2n)!}{2^{2n}(n!)^2}\leq\frac{1}{\sqrt{3n+1}}
\displaystyle \text{for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 37: }\text{Prove that}
\displaystyle 1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\cdots+\frac{1}{n^2}<2-\frac{1}{n}
\displaystyle \text{for all }n\geq2,\ n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n)\text{ be the statement}
\displaystyle 1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\cdots+\frac{1}{n^2}<2-\frac{1}{n}.
\displaystyle \text{For }n=2,
\displaystyle \text{LHS}=1+\frac{1}{2^2}=1+\frac{1}{4}=\frac{5}{4}.
\displaystyle \text{RHS}=2-\frac{1}{2}=\frac{3}{2}.
\displaystyle \text{Since }\frac{5}{4}<\frac{3}{2},\text{ we have LHS}<\text{RHS}.
\displaystyle \therefore P(2)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\geq2,\ k\in N.
\displaystyle \therefore 1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{k^2}<2-\frac{1}{k}.\qquad\text{...(i)}
\displaystyle \text{Now, consider the expression for }n=k+1.
\displaystyle 1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{k^2}+\frac{1}{(k+1)^2}
\displaystyle <2-\frac{1}{k}+\frac{1}{(k+1)^2}\qquad\text{[Using (i)]}
\displaystyle \text{Since }k<k+1,\text{ we have }k(k+1)<(k+1)^2.
\displaystyle \therefore \frac{1}{(k+1)^2}<\frac{1}{k(k+1)}.
\displaystyle \text{Also, }\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}.
\displaystyle \therefore 2-\frac{1}{k}+\frac{1}{(k+1)^2}
\displaystyle <2-\frac{1}{k}+\frac{1}{k(k+1)}
\displaystyle =2-\frac{1}{k}+\frac{1}{k}-\frac{1}{k+1}
\displaystyle =2-\frac{1}{k+1}.
\displaystyle \therefore 1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{k^2}+\frac{1}{(k+1)^2}
\displaystyle <2-\frac{1}{k+1}.
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle 1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\cdots+\frac{1}{n^2}<2-\frac{1}{n}
\displaystyle \text{for all }n\geq2,\ n\in N.
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{Prove that }x^{2n-1}+y^{2n-1}\text{ is divisible by }x+y
\displaystyle \text{for all }n\in N,\text{ where }x,y\in Z.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n)\text{ be the statement that }x^{2n-1}+y^{2n-1}
\displaystyle \text{is divisible by }x+y.
\displaystyle \text{For }n=1,
\displaystyle x^{2(1)-1}+y^{2(1)-1}=x+y.
\displaystyle \text{Clearly, }x+y\text{ is divisible by }x+y.
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore x^{2k-1}+y^{2k-1}\text{ is divisible by }x+y.
\displaystyle \therefore x^{2k-1}+y^{2k-1}=(x+y)\mu
\displaystyle \text{for some integer }\mu.\qquad\text{...(i)}
\displaystyle \text{Now, consider the expression for }n=k+1.
\displaystyle x^{2(k+1)-1}+y^{2(k+1)-1}
\displaystyle =x^{2k+1}+y^{2k+1}
\displaystyle =x^2x^{2k-1}+y^2y^{2k-1}
\displaystyle =x^2\left(x^{2k-1}+y^{2k-1}\right)+(y^2-x^2)y^{2k-1}
\displaystyle =x^2(x+y)\mu-(x^2-y^2)y^{2k-1}\qquad\text{[Using (i)]}
\displaystyle =x^2(x+y)\mu-(x+y)(x-y)y^{2k-1}
\displaystyle =(x+y)\left[x^2\mu-(x-y)y^{2k-1}\right].
\displaystyle \text{Since }x^2\mu-(x-y)y^{2k-1}\text{ is an integer,}
\displaystyle x^{2k+1}+y^{2k+1}\text{ is divisible by }x+y.
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle x^{2n-1}+y^{2n-1}\text{ is divisible by }x+y
\displaystyle \text{for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{Prove that}
\displaystyle \sin x+\sin3x+\cdots+\sin(2n-1)x=\frac{\sin^2nx}{\sin x}
\displaystyle \text{for all }n\in N,\text{ provided }\sin x\neq0.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n)\text{ be the statement}
\displaystyle \sin x+\sin3x+\cdots+\sin(2n-1)x=\frac{\sin^2nx}{\sin x}.
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=\sin x.
\displaystyle \text{RHS}=\frac{\sin^2x}{\sin x}=\sin x.
\displaystyle \therefore \text{LHS}=\text{RHS}.
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore \sin x+\sin3x+\cdots+\sin(2k-1)x
\displaystyle =\frac{\sin^2kx}{\sin x}.\qquad\text{...(i)}
\displaystyle \text{Now, consider the expression for }n=k+1.
\displaystyle \sin x+\sin3x+\cdots+\sin(2k-1)x+\sin(2k+1)x
\displaystyle =\frac{\sin^2kx}{\sin x}+\sin(2k+1)x\qquad\text{[Using (i)]}
\displaystyle =\frac{\sin^2kx+\sin x\sin(2k+1)x}{\sin x}.
\displaystyle \text{Using }2\sin A\sin B=\cos(A-B)-\cos(A+B),
\displaystyle \sin x\sin(2k+1)x
\displaystyle =\frac{1}{2}\left[\cos2kx-\cos2(k+1)x\right].
\displaystyle \text{Also, }\sin^2kx=\frac{1-\cos2kx}{2}.
\displaystyle \therefore \sin^2kx+\sin x\sin(2k+1)x
\displaystyle =\frac{1-\cos2kx}{2}+\frac{\cos2kx-\cos2(k+1)x}{2}
\displaystyle =\frac{1-\cos2(k+1)x}{2}
\displaystyle =\sin^2(k+1)x.
\displaystyle \therefore \sin x+\sin3x+\cdots+\sin(2k+1)x
\displaystyle =\frac{\sin^2(k+1)x}{\sin x}.
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle \sin x+\sin3x+\cdots+\sin(2n-1)x=\frac{\sin^2nx}{\sin x}
\displaystyle \text{for all }n\in N,\text{ provided }\sin x\neq0.
\displaystyle \\

\displaystyle \textbf{Question 40: }\text{Prove that}
\displaystyle \cos\alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+\cdots
\displaystyle +\cos\{\alpha+(n-1)\beta\}
\displaystyle =\frac{\cos\left\{\alpha+\left(\frac{n-1}{2}\right)\beta\right\}\sin\left(\frac{n\beta}{2}\right)}{\sin\left(\frac{\beta}{2}\right)},
\displaystyle \text{for all }n\in N,\text{ provided }\sin\left(\frac{\beta}{2}\right)\neq0.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n)\text{ be the statement}
\displaystyle \cos\alpha+\cos(\alpha+\beta)+\cdots+\cos\{\alpha+(n-1)\beta\}
\displaystyle =\frac{\cos\left\{\alpha+\left(\frac{n-1}{2}\right)\beta\right\}\sin\left(\frac{n\beta}{2}\right)}{\sin\left(\frac{\beta}{2}\right)}.
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=\cos\{\alpha+(1-1)\beta\}=\cos\alpha.
\displaystyle \text{RHS}=\frac{\cos\left\{\alpha+\left(\frac{1-1}{2}\right)\beta\right\}\sin\left(\frac{\beta}{2}\right)}{\sin\left(\frac{\beta}{2}\right)}
\displaystyle =\cos\alpha.
\displaystyle \therefore \text{LHS}=\text{RHS}.
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore \cos\alpha+\cos(\alpha+\beta)+\cdots+\cos\{\alpha+(k-1)\beta\}
\displaystyle =\frac{\cos\left\{\alpha+\left(\frac{k-1}{2}\right)\beta\right\}\sin\left(\frac{k\beta}{2}\right)}{\sin\left(\frac{\beta}{2}\right)}.\qquad\text{...(i)}
\displaystyle \text{Now, consider the expression for }n=k+1.
\displaystyle \cos\alpha+\cos(\alpha+\beta)+\cdots+\cos\{\alpha+(k-1)\beta\}
\displaystyle +\cos(\alpha+k\beta)
\displaystyle =\frac{\cos\left\{\alpha+\left(\frac{k-1}{2}\right)\beta\right\}\sin\left(\frac{k\beta}{2}\right)}{\sin\left(\frac{\beta}{2}\right)}
\displaystyle \quad+\cos(\alpha+k\beta)\qquad\text{[Using (i)]}
\displaystyle =\frac{\sin\left(\frac{k\beta}{2}\right)\cos\left\{\alpha+\left(\frac{k-1}{2}\right)\beta\right\}}{\sin\left(\frac{\beta}{2}\right)}
\displaystyle \quad+\frac{\sin\left(\frac{\beta}{2}\right)\cos(\alpha+k\beta)}{\sin\left(\frac{\beta}{2}\right)}
\displaystyle =\frac{1}{2\sin\left(\frac{\beta}{2}\right)}\left[\sin\left\{\alpha+\left(\frac{2k-1}{2}\right)\beta\right\}\right.
\displaystyle \left.+\sin\left(-\alpha+\frac{\beta}{2}\right)+\sin\left\{\alpha+\left(\frac{2k+1}{2}\right)\beta\right\}\right.
\displaystyle \left.+\sin\left\{-\alpha-\left(\frac{2k-1}{2}\right)\beta\right\}\right].
\displaystyle \text{Since }\sin(-\theta)=-\sin\theta,
\displaystyle \sin\left\{-\alpha-\left(\frac{2k-1}{2}\right)\beta\right\}
\displaystyle =-\sin\left\{\alpha+\left(\frac{2k-1}{2}\right)\beta\right\}.
\displaystyle \therefore \text{the expression becomes}
\displaystyle \frac{1}{2\sin\left(\frac{\beta}{2}\right)}\left[\sin\left\{\alpha+\left(\frac{2k+1}{2}\right)\beta\right\}\right.
\displaystyle \left.+\sin\left(-\alpha+\frac{\beta}{2}\right)\right].
\displaystyle \text{Using }\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2},
\displaystyle =\frac{1}{2\sin\left(\frac{\beta}{2}\right)}\left[2\sin\left(\frac{(k+1)\beta}{2}\right)\right.
\displaystyle \left.\times\cos\left(\alpha+\frac{k\beta}{2}\right)\right]
\displaystyle =\frac{\cos\left(\alpha+\frac{k\beta}{2}\right)\sin\left(\frac{(k+1)\beta}{2}\right)}{\sin\left(\frac{\beta}{2}\right)}
\displaystyle =\frac{\cos\left\{\alpha+\left(\frac{(k+1)-1}{2}\right)\beta\right\}\sin\left(\frac{(k+1)\beta}{2}\right)}{\sin\left(\frac{\beta}{2}\right)}.
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle \cos\alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+\cdots
\displaystyle +\cos\{\alpha+(n-1)\beta\}
\displaystyle =\frac{\cos\left\{\alpha+\left(\frac{n-1}{2}\right)\beta\right\}\sin\left(\frac{n\beta}{2}\right)}{\sin\left(\frac{\beta}{2}\right)}
\displaystyle \text{for all }n\in N,\text{ provided }\sin\left(\frac{\beta}{2}\right)\neq0.
\displaystyle \\

\displaystyle \textbf{Question 41: }\text{Prove that}
\displaystyle \frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}>\frac{13}{24}
\displaystyle \text{for all natural numbers }n>1.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n)\text{ be the statement}
\displaystyle \frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}>\frac{13}{24}.
\displaystyle \text{For }n=2,
\displaystyle \text{LHS}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}=\frac{14}{24}.
\displaystyle \text{RHS}=\frac{13}{24}.
\displaystyle \text{Since }\frac{14}{24}>\frac{13}{24},\text{ we have LHS}>\text{RHS}.
\displaystyle \therefore P(2)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some natural number }k>1.
\displaystyle \therefore \frac{1}{k+1}+\frac{1}{k+2}+\cdots+\frac{1}{2k}>\frac{13}{24}.\qquad\text{...(i)}
\displaystyle \text{Now, consider the expression for }n=k+1.
\displaystyle \frac{1}{k+2}+\frac{1}{k+3}+\cdots+\frac{1}{2k+2}
\displaystyle =\left(\frac{1}{k+1}+\frac{1}{k+2}+\cdots+\frac{1}{2k}\right)
\displaystyle \quad-\frac{1}{k+1}+\frac{1}{2k+1}+\frac{1}{2k+2}
\displaystyle >\frac{13}{24}-\frac{1}{k+1}+\frac{1}{2k+1}+\frac{1}{2k+2}\qquad\text{[Using (i)]}
\displaystyle =\frac{13}{24}-\frac{1}{k+1}+\frac{1}{2k+1}+\frac{1}{2(k+1)}
\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{1}{2(k+1)(2k+1)}
\displaystyle >\frac{13}{24}.
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle \frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}>\frac{13}{24}
\displaystyle \text{for all natural numbers }n>1.
\displaystyle \\

\displaystyle \textbf{Question 42: }\text{Given}
\displaystyle a_1=\frac{1}{2}\left(a_0+\frac{A}{a_0}\right),\quad a_2=\frac{1}{2}\left(a_1+\frac{A}{a_1}\right)
\displaystyle \text{and }a_{n+1}=\frac{1}{2}\left(a_n+\frac{A}{a_n}\right),\text{ where }a_0>0,\ A>0,
\displaystyle \text{prove that, for }n\geq2,
\displaystyle \frac{a_n-\sqrt{A}}{a_n+\sqrt{A}}=\left(\frac{a_1-\sqrt{A}}{a_1+\sqrt{A}}\right)^{2^{n-1}}.
\displaystyle \text{Answer:}
\displaystyle \text{Since }a_0>0\text{ and }A>0,\text{ the recurrence gives }a_n>0\text{ for all }n\in N.
\displaystyle \text{Hence, all the denominators occurring below are non-zero.}
\displaystyle \text{Let }P(n)\text{ be the statement}
\displaystyle \frac{a_n-\sqrt{A}}{a_n+\sqrt{A}}=\left(\frac{a_1-\sqrt{A}}{a_1+\sqrt{A}}\right)^{2^{n-1}}.
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=\frac{a_1-\sqrt{A}}{a_1+\sqrt{A}}.
\displaystyle \text{RHS}=\left(\frac{a_1-\sqrt{A}}{a_1+\sqrt{A}}\right)^{2^{1-1}}
\displaystyle =\left(\frac{a_1-\sqrt{A}}{a_1+\sqrt{A}}\right)^1
\displaystyle =\frac{a_1-\sqrt{A}}{a_1+\sqrt{A}}.
\displaystyle \therefore \text{LHS}=\text{RHS}.
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\geq1.
\displaystyle \therefore \frac{a_k-\sqrt{A}}{a_k+\sqrt{A}}
\displaystyle =\left(\frac{a_1-\sqrt{A}}{a_1+\sqrt{A}}\right)^{2^{k-1}}.\qquad\text{...(i)}
\displaystyle \text{Now, consider the expression for }n=k+1.
\displaystyle \frac{a_{k+1}-\sqrt{A}}{a_{k+1}+\sqrt{A}}
\displaystyle =\frac{\frac{1}{2}\left(a_k+\frac{A}{a_k}\right)-\sqrt{A}}{\frac{1}{2}\left(a_k+\frac{A}{a_k}\right)+\sqrt{A}}
\displaystyle =\frac{a_k+\frac{A}{a_k}-2\sqrt{A}}{a_k+\frac{A}{a_k}+2\sqrt{A}}
\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 =\left(\frac{a_k-\sqrt{A}}{a_k+\sqrt{A}}\right)^2
\displaystyle =\left[\left(\frac{a_1-\sqrt{A}}{a_1+\sqrt{A}}\right)^{2^{k-1}}\right]^2\qquad\text{[Using (i)]}
\displaystyle =\left(\frac{a_1-\sqrt{A}}{a_1+\sqrt{A}}\right)^{2\cdot2^{k-1}}
\displaystyle =\left(\frac{a_1-\sqrt{A}}{a_1+\sqrt{A}}\right)^{2^k}
\displaystyle =\left(\frac{a_1-\sqrt{A}}{a_1+\sqrt{A}}\right)^{2^{(k+1)-1}}.
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle \frac{a_n-\sqrt{A}}{a_n+\sqrt{A}}
\displaystyle =\left(\frac{a_1-\sqrt{A}}{a_1+\sqrt{A}}\right)^{2^{n-1}}
\displaystyle \text{for all }n\geq1,\text{ and hence for all }n\geq2.
\displaystyle \\

\displaystyle \textbf{Question 43: }\text{Let }P(n)\text{ be the statement }2^n\geq3n.
\displaystyle \text{If }P(r)\text{ is true, show that }P(r+1)\text{ is true. Do you conclude that}
\displaystyle P(n)\text{ is true for all }n\in N?
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(r)\text{ be true for some }r\in N.
\displaystyle \therefore 2^r\geq3r.
\displaystyle \text{Multiplying both sides by }2,
\displaystyle 2^{r+1}\geq6r.
\displaystyle \text{Since }r\geq1,
\displaystyle 6r-3(r+1)=3r-3=3(r-1)\geq0.
\displaystyle \therefore 6r\geq3(r+1).
\displaystyle \therefore 2^{r+1}\geq3(r+1).
\displaystyle \therefore P(r+1)\text{ is true.}
\displaystyle \text{However, this does not imply that }P(n)\text{ is true for all }n\in N.
\displaystyle \text{For }n=1,
\displaystyle 2^1=2<3=3(1).
\displaystyle \therefore P(1)\text{ is false.}
\displaystyle \text{Also, }2^2=4<6\text{ and }2^3=8<9.
\displaystyle \therefore P(2)\text{ and }P(3)\text{ are also false.}
\displaystyle \text{Hence, }P(n)\text{ is not true for all }n\in N.
\displaystyle \text{However, }P(4)\text{ is true since }2^4=16\geq12.
\displaystyle \text{Therefore, using }P(r)\Rightarrow P(r+1),\text{ we conclude that}
\displaystyle 2^n\geq3n\text{ for all }n\geq4.
\displaystyle \\

\displaystyle \textbf{Question 44: }\text{Show by the principle of mathematical induction that the sum }S_n
\displaystyle \text{of the first }n\text{ terms of the series}
\displaystyle 1^2+2\times2^2+3^2+2\times4^2+5^2+2\times6^2+7^2+\cdots
\displaystyle \text{is given by}
\displaystyle S_n=\begin{cases}\dfrac{n(n+1)^2}{2},&\text{if }n\text{ is even},\\[4pt]\dfrac{n^2(n+1)}{2},&\text{if }n\text{ is odd}.\end{cases}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n)\text{ be the statement}
\displaystyle S_n=\begin{cases}\dfrac{n(n+1)^2}{2},&\text{if }n\text{ is even},\\[4pt]\dfrac{n^2(n+1)}{2},&\text{if }n\text{ is odd}.\end{cases}
\displaystyle \text{For }n=1,\text{ which is odd,}
\displaystyle \text{LHS}=S_1=1^2=1.
\displaystyle \text{RHS}=\frac{1^2(1+1)}{2}=1.
\displaystyle \therefore \text{LHS}=\text{RHS}.
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore S_k=\begin{cases}\dfrac{k(k+1)^2}{2},&\text{if }k\text{ is even},\\[4pt]\dfrac{k^2(k+1)}{2},&\text{if }k\text{ is odd}.\end{cases}\qquad\text{...(i)}
\displaystyle \text{Now, we prove that }P(k+1)\text{ is true.}
\displaystyle \text{Case I: Let }k\text{ be even.}
\displaystyle \text{Then }k+1\text{ is odd, and the }(k+1)\text{th term is }(k+1)^2.
\displaystyle \therefore S_{k+1}=S_k+(k+1)^2
\displaystyle =\frac{k(k+1)^2}{2}+(k+1)^2\qquad\text{[Using (i)]}
\displaystyle =(k+1)^2\left(\frac{k}{2}+1\right)
\displaystyle =\frac{(k+1)^2(k+2)}{2}
\displaystyle =\frac{(k+1)^2\{(k+1)+1\}}{2}.
\displaystyle \text{Thus, the required formula holds when }k+1\text{ is odd.}
\displaystyle \text{Case II: Let }k\text{ be odd.}
\displaystyle \text{Then }k+1\text{ is even, and the }(k+1)\text{th term is }2(k+1)^2.
\displaystyle \therefore S_{k+1}=S_k+2(k+1)^2
\displaystyle =\frac{k^2(k+1)}{2}+2(k+1)^2\qquad\text{[Using (i)]}
\displaystyle =\frac{k+1}{2}\left[k^2+4(k+1)\right]
\displaystyle =\frac{k+1}{2}\left(k^2+4k+4\right)
\displaystyle =\frac{(k+1)(k+2)^2}{2}
\displaystyle =\frac{(k+1)\{(k+1)+1\}^2}{2}.
\displaystyle \text{Thus, the required formula holds when }k+1\text{ is even.}
\displaystyle \therefore P(k+1)\text{ is true in both cases.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle S_n=\begin{cases}\dfrac{n(n+1)^2}{2},&\text{if }n\text{ is even},\\[4pt]\dfrac{n^2(n+1)}{2},&\text{if }n\text{ is odd},\end{cases}
\displaystyle \text{for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 45: }\text{Prove that the number of subsets of a set containing }n
\displaystyle \text{distinct elements is }2^n\text{ for all }n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n)\text{ be the statement that a set containing }n\text{ distinct elements}
\displaystyle \text{has }2^n\text{ subsets.}
\displaystyle \text{For }n=1,\text{ let the set be }\{a\}.
\displaystyle \text{Its subsets are }\phi\text{ and }\{a\}.
\displaystyle \therefore \text{The number of subsets}=2=2^1.
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore \text{a set containing }k\text{ distinct elements has }2^k\text{ subsets.}\qquad\text{...(i)}
\displaystyle \text{Now, let }A=\{a_1,a_2,\ldots,a_k,a_{k+1}\}.
\displaystyle \text{Let }B=\{a_1,a_2,\ldots,a_k\}.
\displaystyle \text{Then }B\text{ contains }k\text{ distinct elements.}
\displaystyle \text{By the induction hypothesis, }B\text{ has }2^k\text{ subsets.}
\displaystyle \text{The subsets of }A\text{ can be divided into two groups.}
\displaystyle \text{The first group consists of subsets not containing }a_{k+1}.
\displaystyle \text{These are precisely the subsets of }B,\text{ so their number is }2^k.
\displaystyle \text{The second group consists of subsets containing }a_{k+1}.
\displaystyle \text{Each such subset is obtained by adjoining }a_{k+1}\text{ to a subset of }B.
\displaystyle \text{Hence, the number of such subsets is also }2^k.
\displaystyle \therefore \text{The total number of subsets of }A
\displaystyle =2^k+2^k=2\cdot2^k=2^{k+1}.
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle \text{the number of subsets of a set containing }n\text{ distinct elements is }2^n
\displaystyle \text{for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 46: }\text{A sequence }a_1,a_2,a_3,\ldots\text{ is defined by }a_1=3
\displaystyle \text{and }a_k=7a_{k-1}\text{ for all natural numbers }k\geq2.
\displaystyle \text{Show that }a_n=3\cdot7^{n-1}\text{ for all }n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n)\text{ be the statement }a_n=3\cdot7^{n-1}.
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=a_1=3.
\displaystyle \text{RHS}=3\cdot7^{1-1}=3\cdot7^0=3.
\displaystyle \therefore \text{LHS}=\text{RHS}.
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore a_k=3\cdot7^{k-1}.\qquad\text{...(i)}
\displaystyle \text{Now, consider }a_{k+1}.
\displaystyle a_{k+1}=7a_k
\displaystyle =7\left(3\cdot7^{k-1}\right)\qquad\text{[Using (i)]}
\displaystyle =3\cdot7^k
\displaystyle =3\cdot7^{(k+1)-1}.
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle a_n=3\cdot7^{n-1}\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 47: }\text{A sequence }x_1,x_2,x_3,\ldots\text{ is defined by }x_1=2
\displaystyle \text{and }x_k=\frac{x_{k-1}}{k}\text{ for all natural numbers }k\geq2.
\displaystyle \text{Show that }x_n=\frac{2}{n!}\text{ for all }n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n)\text{ be the statement }x_n=\frac{2}{n!}.
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=x_1=2.
\displaystyle \text{RHS}=\frac{2}{1!}=2.
\displaystyle \therefore \text{LHS}=\text{RHS}.
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore x_k=\frac{2}{k!}.\qquad\text{...(i)}
\displaystyle \text{Now, consider }x_{k+1}.
\displaystyle x_{k+1}=\frac{x_k}{k+1}
\displaystyle =\frac{1}{k+1}\cdot\frac{2}{k!}\qquad\text{[Using (i)]}
\displaystyle =\frac{2}{(k+1)k!}
\displaystyle =\frac{2}{(k+1)!}.
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle x_n=\frac{2}{n!}\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 48: }\text{A sequence }x_0,x_1,x_2,x_3,\ldots\text{ is defined by }x_0=5
\displaystyle \text{and }x_k=4+x_{k-1}\text{ for all natural numbers }k\geq1.
\displaystyle \text{Show by mathematical induction that }x_n=5+4n\text{ for all }n\geq0.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n)\text{ be the statement }x_n=5+4n.
\displaystyle \text{For }n=0,
\displaystyle \text{LHS}=x_0=5.
\displaystyle \text{RHS}=5+4(0)=5.
\displaystyle \therefore \text{LHS}=\text{RHS}.
\displaystyle \therefore P(0)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some non-negative integer }k.
\displaystyle \therefore x_k=5+4k.\qquad\text{...(i)}
\displaystyle \text{Now, consider }x_{k+1}.
\displaystyle x_{k+1}=4+x_k
\displaystyle =4+(5+4k)\qquad\text{[Using (i)]}
\displaystyle =5+4k+4
\displaystyle =5+4(k+1).
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle x_n=5+4n\text{ for all non-negative integers }n.
\displaystyle \\

\displaystyle \textbf{Question 49: }\text{Using the principle of mathematical induction, prove that}
\displaystyle \sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}
\displaystyle \text{for all natural numbers }n\geq2.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n)\text{ be the statement}
\displaystyle \sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}.
\displaystyle \text{For }n=2,
\displaystyle \text{LHS}=\sqrt{2}.
\displaystyle \text{RHS}=1+\frac{1}{\sqrt{2}}.
\displaystyle \text{Now, }1+\frac{1}{\sqrt{2}}-\sqrt{2}
\displaystyle =1-\frac{1}{\sqrt{2}}>0.
\displaystyle \therefore \sqrt{2}<1+\frac{1}{\sqrt{2}}.
\displaystyle \therefore P(2)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some natural number }k\geq2.
\displaystyle \therefore \sqrt{k}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{k}}.\qquad\text{...(i)}
\displaystyle \text{Now, consider the expression for }n=k+1.
\displaystyle \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}
\displaystyle >\sqrt{k}+\frac{1}{\sqrt{k+1}}\qquad\text{[Using (i)]}
\displaystyle \text{Also,}
\displaystyle \sqrt{k+1}-\sqrt{k}
\displaystyle =\frac{(k+1)-k}{\sqrt{k+1}+\sqrt{k}}
\displaystyle =\frac{1}{\sqrt{k+1}+\sqrt{k}}
\displaystyle <\frac{1}{\sqrt{k+1}}.
\displaystyle \therefore \sqrt{k+1}<\sqrt{k}+\frac{1}{\sqrt{k+1}}.
\displaystyle \therefore \sqrt{k+1}
\displaystyle <\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}.
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle \sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}
\displaystyle \text{for all natural numbers }n\geq2.
\displaystyle \\

\displaystyle \textbf{Question 50: }\text{The distributive law states that, for all real numbers }c,a_1,a_2,
\displaystyle c(a_1+a_2)=ca_1+ca_2.\text{ Using this law and mathematical induction, prove that}
\displaystyle \text{for every natural number }n\geq2\text{ and all real numbers }c,a_1,a_2,\ldots,a_n,
\displaystyle c(a_1+a_2+\cdots+a_n)=ca_1+ca_2+\cdots+ca_n.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n)\text{ be the statement}
\displaystyle c(a_1+a_2+\cdots+a_n)=ca_1+ca_2+\cdots+ca_n,
\displaystyle \text{where }c,a_1,a_2,\ldots,a_n\in\mathbb{R}.
\displaystyle \text{For }n=2,
\displaystyle \text{LHS}=c(a_1+a_2).
\displaystyle \text{By the distributive law,}
\displaystyle c(a_1+a_2)=ca_1+ca_2=\text{RHS}.
\displaystyle \therefore P(2)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some natural number }k\geq2.
\displaystyle \therefore c(a_1+a_2+\cdots+a_k)=ca_1+ca_2+\cdots+ca_k.\qquad\text{...(i)}
\displaystyle \text{Now, consider the expression for }n=k+1.
\displaystyle c(a_1+a_2+\cdots+a_k+a_{k+1})
\displaystyle =c\left[(a_1+a_2+\cdots+a_k)+a_{k+1}\right]
\displaystyle =c(a_1+a_2+\cdots+a_k)+ca_{k+1}
\displaystyle \qquad\text{[By the distributive law]}
\displaystyle =ca_1+ca_2+\cdots+ca_k+ca_{k+1}\qquad\text{[Using (i)]}
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle c(a_1+a_2+\cdots+a_n)=ca_1+ca_2+\cdots+ca_n
\displaystyle \text{for every natural number }n\geq2\text{ and all }c,a_1,a_2,\ldots,a_n\in\mathbb{R}.
\displaystyle \\

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