NCERT Solutions Class 11 Maths Chapter-4 (Principle of Mathematical Induction)

NCERT Solutions Class 11 Maths from class 11th Students will get the answers of Chapter-4 (Principle of Mathematical Induction) . This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.

We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

NCERT Question-Answer

Class 11 Mathematics

Chapter-4 (Principle of Mathematical Induction)

Questions and answers given in practice

Chapter-4 (Principle of Mathematical Induction)

Exercise 4.1

Q1. Prove the following by using the principle of mathematical induction for all n ∈ N: $1+3+3^2+\dots +3^{a-1}=\frac{(3^n-1)}{2}$

Answer. $\begin{array}{c}\text{Let the given statement be}P\left(n\right),\text{i.e.,}\\ P\left(n\right):1+3+3^2+\dots +3^{n-1}=\frac{(3^n-1)}{2}\\ \text{For}n=1,\text{we have}\\ P\left(1\right):1=\frac{(3^{\prime }-1)}{2}=\frac{3-1}{2}=\frac{2}{2}=1\end{array}$ , which is true $\begin{array}{c}\text{Let}P\left(k\right)\text{be true for some positive integer}k,\text{i.e.,}\\ 1+3+3^2+\dots +3^{k-1}=\frac{(3^k-1)}{2}\\ \text{We shall now prove that}P(k+1)\text{is true.}\\ \text{Consider}\\ 1+3+3^2+\dots +3^{k-1}+3^{(k+1)-1}\\ =(1+3+3^2+\dots +3^{k-1})+3^k\end{array}$ $\begin{array}{c}=\frac{(3^t-1)}{2}+3^k\\ =\frac{(3^t-1)+{2.3}^k}{2}\end{array}$ [using (i)] $=\frac{(1+2)3^k-1}{2}$ $\begin{array}{cc}& =\frac{{3.3}^k-1}{2}\\ & =\frac{3^{k+1}-1}{2}\end{array}$ $\begin{array}{c}\text{Thus,}P(k+1)\text{is true whenever}P\left(k\right)\text{is true.}\\ \text{Hence, by the principle of mathematical induction, statement}P\left(n\right)\text{is true for all natural}\\ \text{numbers l.e.,}\end{array}$

Q2. Prove the following by using the principle of mathematical induction for all n ∈ N: $1^3+2^3+3^3+\dots +n^3={\left(\frac{n(n+1)}{2}\right)}^2$

Answer. $\begin{array}{c}\text{Let the given statement be}P\left(n\right),\text{i.e.,}\\ 1^3+2^3+3^3+\dots +n^3={\left(\frac{n(n+1)}{2}\right)}^2\\ \text{For}n=1,\text{we have}\\ \text{For}n=1,\text{we have}\\ P\left(1\right):1^3=1={(\begin{array}{c}1(1+1)\end{array}}^2={\left(\frac{1.2}{2}\right)}^2=1^2=1\end{array}$ , which is true. $\begin{array}{c}\text{Let}P\left(k\right)\text{be true for some positive integer}k,\text{l.e.,}\\ 1^3+2^3+3^3+\dots \dots +k^3={\left(\frac{k(k+1)}{2}\right)}^2\\ \text{We shall now prove that}P(k+1)\text{is true.}\\ \text{Consider}\\ 1^3+2^3+3^3+\dots +k^3+(k+1{)}^3\end{array}$ $={\left(\frac{k(k+1)}{2}\right)}^2+(k+1{)}^3[\text{Using}\left(i\right)]$ $\begin{array}{cc}& =\frac{k^2(k+1{)}^2}{4}+(k+1{)}^3\\ & =\frac{k^2(k+1{)}^2+4(k+1{)}^3}{4}\\ & =\frac{(k+1{)}^2\{k^2+4(k+1\left)\right\}}{4}\end{array}$ $\begin{array}{c}=\frac{(k+1{)}^2\{k^2+4k+4\}}{4}\\ =\frac{(k+1{)}^2(k+2{)}^2}{4}\\ =\frac{(k+1{)}^2(k+1+1{)}^2}{4}\end{array}$ $=(1^3+2^3+3^3+\dots .+k^3)+(k+1{)}^3={\left(\frac{(k+1)(k+1+1)}{2}\right)}^2$ $\begin{array}{c}\text{Thus,}P(k+1)\text{is true whenever}P\left(k\right)\text{is true.}\\ \text{Hence, by the principle of mathematical induction, statement}P\left(n\right)\text{is true for all natural}\\ \text{numbers l.e.,}\end{array}$

Q3. Prove the following by using the principle of mathematical induction for all n ∈ N: $1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+\dots +\frac{1}{(1+2+3+\dots n)}=\frac{2n}{(n+1)}$

Answer. $\begin{array}{c}\text{Let the given statement be}P\left(n\right),\text{i.e.,}\\ \text{P(n):}1+\frac{1}{1+2}+\frac{1}{1+2+3}+\dots +\frac{1}{1+2+3+\dots n}=\frac{2n}{n+1}\\ \text{For}n=1,\text{we have}\\ P\left(1\right):1=\frac{2.1}{1+1}=\frac{2}{2}=1\\ \text{Let}P\left(k\right)\text{be true for some positive integer}k,\text{i.e.,}\end{array}$ $1+\frac{1}{1+2}+\dots +\frac{1}{1+2+3}+\dots +\frac{1}{1+2+3+\dots +k}=\frac{2k}{k+1}$............(i) $\begin{array}{c}\text{We shall now prove that}P(k+1)\text{is true.}\\ \text{Consider}\\ 1+\frac{1}{1+2}+\frac{1}{1+2+3}+\dots +\frac{1}{1+2+3+\dots +k}+\frac{1}{1+2+2+k+(k+1)}\end{array}$ $\begin{array}{c}=(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\dots +\frac{1}{1+2+3+.k})+\frac{1}{1+2+3+\dots +k+(k+1)}\\ =\frac{2k}{k+1}+\frac{1}{1+2+3+\dots +k+(k+1)}\left[\text{Using (i)}\right]\end{array}$ $=\frac{2k}{k+1}+\frac{1}{\left(\frac{(k+1)(k+1+1)}{2}\right)}$ $[1+2+3+\dots +n=\frac{n(n+1)}{2}]$ $\begin{array}{c}=\frac{2k}{(k+1)}+\frac{2}{(k+1)(k+2)}\\ =\frac{2}{(k+1)}(k+\frac{1}{k+2})\end{array}$

Q4. Prove the following by using the principle of mathematical induction for all n ∈ N: $\mathrm{1.2.3}+2.3.4+\dots +n(n+1)(n+2)=\frac{n(n+1)(n+2)(n+3)}{4}$

Answer. $\begin{array}{c}\text{Let the given statement be}P\left(n\right),\text{i.e.,}\\ P\left(n\right):1.2.3+2.3.4+\dots +n(n+1)(n+2)=\frac{n(n+1)(n+2)(n+3)}{4}\end{array}$ $\begin{array}{c}\text{For}n=1,\text{we have}\\ P\left(1\right):1.2\cdot 3=6=\frac{1(1+1)(1+2)(1+3)}{4}=\frac{1.23.4}{4}=6\end{array}$ , which is true Let P(k) be true for some positive integer k, i.e., $1.2.3+2.3.4+\dots +k(k+1)(k+2)=\frac{k(k+1)(k+2)(k+3)}{4}$ ……………(i) $\begin{array}{c}\text{We shall now prove that}P(k+1)\text{is true.}\\ \text{Consider}\\ 1.2.3+2.3.4+\dots +k(k+1)(k+2)+(k+1)(k+2)(k+3)\\ =\{1.2.3+2.3.4+\dots +k(k+1\left)\right(k+2\left)\right\}+(k+1)(k+2)(k+3)\end{array}$ $\begin{array}{c}=\frac{k(k+1)(k+2)(k+3)}{4}+(k+1)(k+2)(k+3)\left[\text{Using}\right(i\left)\right]\\ =(k+1)(k+2)(k+3)(\frac{k}{4}+1)\\ =\frac{(k+1)(k+2)(k+3)(k+4)}{4}\\ =\frac{(k+1)(k+1+1)(k+1+2)(k+1+3)}{4}\end{array}$ $\begin{array}{c}\text{Thus,}P(k+1)\text{is true whenever}P\left(k\right)\text{is true.}\\ \text{Hence, by the principle of mathematical induction, statement}P\left(n\right)\text{is true for all natural}\\ \text{numbers l.e.,}n\text{.}\end{array}$

Q5. Prove the following by using the principle of mathematical induction for all n ∈ N: $1.3+{2.3}^2+{3.3}^3+\dots +n{3}^n=\frac{(2n-1)3^{n+1}+3}{4}$

Answer. $\begin{array}{c}\text{Let the given statement be}P\left(n\right),\text{i.e.,}\\ 1.3+{2.3}^2+{3.3}^{\circ }+\dots +n{3}^n=\frac{(2n-1)3^{n+1}+3}{4}\end{array}$ $\begin{array}{c}P\left(n\right):\\ \text{For}n=1,\text{we have}\end{array}$ $P\left(1\right):1.3=3=\frac{(2.1-1)3^{1+1}+3}{4}=\frac{3^2+3}{4}=\frac{12}{4}=3$ $\begin{array}{c}\text{Let}P\left(k\right)\text{be true for some positive integer}k,\text{i.e.,}\\ 1.3+{2.3}^2+{3.3}^3+\dots +k{3}^4=\frac{(2k-1)3^{k+1}+3}{4}\\ \text{We shall now prove that}P(k+1)\text{is true.}\end{array}$ ……….(i) Consider $\begin{array}{c}1.3+{2.3}^2+{3.3}^3+\dots +k{3}^k+(k+1)3^{k+1}\\ =(1.3+{2.3}^2+{3.3}^3+\dots +k{.3}^k)+(k+1)3^{k+1}\end{array}$ $=\frac{(2k-1)3^{k+1}+3}{4}+(k+1)3^{k+1}\left[\text{Using}\right(i\left)\right]$ $\begin{array}{c}=\frac{(2k-1)3^{3+1}+3+4(k+1)3^{k+1}}{4}\\ =\frac{3^{t+1}\{2k-1+4(k+1\left)\right\}+3}{4}\end{array}$ $\begin{array}{cc}& =\frac{3^{k+1}\{6k+3\}+3}{4}\\ & =\frac{3^{k+1}\cdot 3\{2k+1\}+3}{4}\end{array}$ $\begin{array}{c}=\frac{3^{(k+1)+1}\{2k+1\}+3}{4}\\ =\frac{\left\{2\right(k+1)-1\}{3}^{(k+1)+1}+3}{4}\end{array}$ Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Q6. Prove the following by using the principle of mathematical induction for all n ∈ N: $1.2+2.3+3.4+\dots +n.(n+1)=\left[\frac{n(n+1)(n+2)}{3}\right]$

Answer. Let the given statement be P(n), i.e., P(n) : $1.2+2.3+3.4+\dots +n\cdot (n+1)=\left[\frac{n(n+1)(n+2)}{3}\right]$ For n = 1, we have P(1): $1.2=2=\frac{1(1+1)(1+2)}{3}=\frac{1.2.3}{3}=2$which is true. Let P(k) be true for some positive integer k, i.e., $1.2+2.3+3.4+\dots .+k.(k+1)=\left[\frac{k(k+1)(k+2)}{3}\right]$ ……….(i) We shall now prove that P(k + 1) is true. Consider 1.2 + 2.3 + 3.4 + … + k.(k + 1) + (k + 1).(k + 2) = [1.2 + 2.3 + 3.4 + … + k.(k + 1)] + (k + 1).(k + 2) $=\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)\left[\text{Using}\right(i\left)\right]$ $\begin{array}{c}=(k+1)(k+2)(\frac{k}{3}+1)\\ =\frac{(k+1)(k+2)(k+3)}{3}\\ =\frac{(k+1)(k+1+1)(k+1+2)}{3}\end{array}$ Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Q7. Prove the following by using the principle of mathematical induction for all n ∈ N: $1.3+3.5+5.7+\dots +(2n-1)(2n+1)=\frac{n(4n^2+6n-1)}{3}$

Answer. Let the given statement be P(n), i.e., P(n) : $1.3+3.5+5.7+\dots +(2n-1)(2n+1)=\frac{n(4n^2+6n-1)}{3}$ $\begin{array}{c}\text{For}n=1,\text{we have}\\ P\left(1\right):1.3=3=\frac{1({4.1}^2+6.1-1)}{3}=\frac{4+6-1}{3}=\frac{9}{3}=3\end{array}$ , which is true Let P(k) be true for sorne positive integer k, i.e., $1.3+3.5+5.7+\dots .+(2k-1)(2k+1)=\frac{k(4k^2+6k-1)}{3}$ ……..(i) $\begin{array}{c}\text{We shall now prove that}P(k+1)\text{is true.}\\ \text{Consider}\\ (1.3+3.5+5.7+\dots +(2k-1\left)\right(2k+1)+\{2(k+1)-1\left\}\right\{2(k+1)+1\}\end{array}$ $=\frac{k(4k^2+6k-1)}{3}+(2k+2-1)(2k+2+1)\left[\text{Using (i)}\right]$ $\begin{array}{c}=\frac{k(4k^2+6k-1)}{3}+(2k+1)(2k+3)\\ =\frac{k(4k^2+6k-1)}{3}+(4k^2+8k+3)\\ =\frac{k(4k^2+6k-1)+3(4k^2+8k+3)}{3}\\ =\frac{4k^3+6k^2-k+12k^2+24k+9}{3}\\ =\frac{4k^3+18k^2+93k+9}{3}\\ =\frac{4k^3+14k^2+9k+4k^2+14k+9}{3}\end{array}$ $\begin{array}{c}=\frac{k(4k^2+14k+9)+1(4k^2+14k+9)}{3}\\ =\frac{(k+1)(4k^2+14k+9)}{3}\end{array}$ $\begin{array}{c}=\frac{(k+1)\{4k^2+8k+4+6k+6-1\}}{3}\\ =\frac{(k+1)\{4(k^2+2k+1)+6(k+1)-1\}}{3}\\ =\frac{(k+1)\left\{4\right(k+1{)}^2+6(k+1)-1\}}{3}\end{array}$ Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Q8. Prove the following by using the principle of mathematical induction for all n ∈ N: $1.2+{2.2}^2+{3.2}^3+\dots +n{.2}^{\circ }=(n-1)2^{n+1}+2$

Answer. Let the given statement be P(n), i.e., $P\left(n\right):1.2+{2.2}^2+{3.2}^2+\dots +n{.2}^n=(n-1)2^{n+1}+2$ For n = 1, we have $P\left(1\right):1.2=2=(1-1)2^{1+1}+2=0+2=2,$ Let P(k) be true for some positive integer k, i.e., $1.2+{2.2}^2+{3.2}^2+\dots +k{.2}^k=(k-1)2^{k+1}+2\dots$..(i) We shall now prove that P(k + 1) is true. Consider $\begin{array}{c}\{1.2+{2.2}^2+{3.2}^3+\dots +k{.2}^k\}+(k+1)\cdot 2^{k+1}\\ =(k-1)2^{k+1}+2+(k+1)2^{k+1}\\ =2^{k+1}\left\{\right(k-1)+(k+1\left)\right\}+2\\ =2^{k+1}\cdot 2k+2\\ =k{.2}^{(k+1)+1}+2\\ =\left\{\right(k+1)-1\}{2}^{(k+1)+1}+2\end{array}$ Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Q9. Prove the following by using the principle of mathematical induction for all n ∈ N: $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots +\frac{1}{2^n}=1-\frac{1}{2^n}$

Answer. Let the given statement be P(n), i.e., P(n): $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots +\frac{1}{2^n}=1-\frac{1}{2^n}$ For n = 1, we have P(1): $\frac{1}{2}=1-\frac{1}{2^1}=\frac{1}{2}$, which is true Let P(k) be true for some positive integer k, i.e., $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots +\frac{1}{2^k}=1-\frac{1}{2^k}$ We shall now prove that P(k + 1) is true. Consider $\begin{array}{c}(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots \dots +\frac{1}{2^k})+\frac{1}{2^{k+1}}\\ =(1-\frac{1}{2^k})+\frac{1}{2^{k+1}}\end{array}$ [using(i))] $\begin{array}{c}=1-\frac{1}{2^k}+\frac{1}{2\cdot 2^k}\\ =1-\frac{1}{2^k}(1-\frac{1}{2})\\ =1-\frac{1}{2^k}\left(\frac{1}{2}\right)\\ =1-\frac{1}{2^{k+1}}\end{array}$ Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Q10. Prove the following by using the principle of mathematical induction for all n ∈ N: $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\dots +\frac{1}{(3n-1)(3n+2)}=\frac{n}{(6n+4)}$

Answer. Let the given statement be P(n), i.e., P(n): $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\dots +\frac{1}{(3n-1)(3n+2)}=\frac{n}{(6n+4)}$ For n = 1, we have $P\left(1\right)=\frac{1}{2.5}=\frac{1}{10}=\frac{1}{6.1+4}=\frac{1}{10}$ , which is true. Let P(k) be true for some positive integer k, i.e., $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\dots +\frac{1}{(3k-1)(3k+2)}=\frac{k}{6k+4}$ ….(i) We shall now prove that P(k + 1) is true. Consider $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\dots \dots +\frac{1}{(3k-1)(3k+2)}+\frac{1}{\left\{3\right(k+1)-1\}\left\{3\right(k+1)+2\}}$ $=\frac{k}{6k+4}+\frac{1}{(3k+3-1)(3k+3+2)}\left[\text{Using}\right(i\left)\right]$ $\begin{array}{c}=\frac{k}{6k+4}+\frac{1}{(3k+2)(3k+5)}\\ =\frac{k}{2(3k+2)}+\frac{1}{(3k+2)(3k+5)}\\ =\frac{1}{(3k+2)}(\frac{k}{2}+\frac{1}{3k+5})\\ =\frac{1}{(3k+2)}\left(\frac{k(3k+5)+2}{2(3k+5)}\right)\end{array}$ $\begin{array}{c}=\frac{1}{(3k+2)}\left(\frac{3k^2+5k+2}{2(3k+5)}\right)\\ =\frac{1}{(3k+2)}\left(\frac{(3k+2)(k+1)}{2(3k+5)}\right)\\ =\frac{(k+1)}{6k+10}\\ =\frac{(k+1)}{6(k+1)+4}\end{array}$ Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Q11. Prove the following by using the principle of mathematical induction for all n ∈ N: $\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+\frac{1}{\mathrm{3.4.5}}+\dots +\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$

Answer. Let the given statement be P(n),i.e., P(n): $\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+\frac{1}{\mathrm{3.4.5}}+\dots +\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$ For n = 1, we have $P\left(1\right):\frac{1}{1\cdot 2\cdot 3}=\frac{1\cdot (1+3)}{4(1+1)(1+2)}=\frac{1\cdot 4}{4\cdot 2\cdot 3}=\frac{1}{1\cdot 2\cdot 3}$, which is true Let P(k) be true for some positive integer k, i.e., $\frac{1}{1\cdot 2\cdot 3}+\frac{1}{2\cdot 3\cdot 4}+\frac{1}{3\cdot 4\cdot 5}+\dots +\frac{1}{k(k+1)(k+2)}=\frac{k(k+3)}{4(k+1)(k+2)}$ …..(i) We shall now prove that P(k + 1) is true. Consider $[\frac{1}{1\cdot 2\cdot 3}+\frac{1}{2\cdot 3\cdot 4}+\frac{1}{3\cdot 4\cdot 5}+\dots +\frac{1}{k(k+1)(k+2)}]+\frac{1}{(k+1)(k+2)(k+3)}$ $=\frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)}\left[\text{Using}\right(i\left)\right]$ $\begin{array}{c}=\frac{1}{(k+1)(k+2)}\{\frac{k(k+3)}{4}+\frac{1}{k+3}\}\\ =\frac{1}{(k+1)(k+2)}\left\{\frac{k(k+3{)}^2+4}{4(k+3)}\right\}\\ =\frac{1}{(k+1)(k+2)}\left\{\frac{k(k^2+6k+9)+4}{4(k+3)}\right\}\end{array}$ $\begin{array}{c}=\frac{1}{(k+1)(k+2)}\left\{\frac{k^3+6k^2+9k+4}{4(k+3)}\right\}\\ =\frac{1}{(k+1)(k+2)}\left\{\frac{k^3+2k^2+k+4k^2+8k+4}{4(k+3)}\right\}\\ =\frac{1}{(k+1)(k+2)}\left\{\frac{k(k^2+2k+1)+4(k^2+2k+1)}{4(k+3)}\right\}\end{array}$ $\begin{array}{c}=\frac{1}{(k+1)(k+2)}\left\{\frac{k(k+1{)}^2+4(k+1{)}^2}{4(k+3)}\right\}\\ =\frac{(k+1{)}^2(k+4)}{4(k+1)(k+2)(k+3)}\\ =\frac{(k+1)\left\{\right(k+1)+3\}}{4\left\{\right(k+1)+1\}\left\{\right(k+1)+2\}}\end{array}$ Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Q12. Prove the following by using the principle of mathematical induction for all n ∈ N: $a+ar+a{r}^2+\dots +a{r}^{n-1}=\frac{a(r^n-1)}{r-1}$

Answer. Let the given statement be P(n), i.e., P(n): $a+ar+a{r}^2+\dots +a{r}^{n-1}=\frac{a(r^n-1)}{r-1}$ For n = 1, we have $P\left(1\right):a=\frac{a(r^1-1)}{(r-1)}=a$, which is true. Let P(k) be true for some positive integer k, i.e., $a+ar+a{r}^2+\dots \dots .+a{r}^{k-1}=\frac{a(r^k-1)}{r-1}$ …….(i) We shall now prove that P(k + 1) is true. Consider $\begin{array}{c}\{a+ar+a{r}^2+\dots \dots +a{r}^{k-1}\}+a{r}^{(k+1)-1}\\ =\frac{a(r^k-1)}{r-1}+a{r}^k\end{array}$ [ Using(i) ] $\begin{array}{cc}& =\frac{a(r^k-1)+a{r}^k(r-1)}{r-1}\\ & =\frac{a(r^k-1)+a{r}^{k+1}-a{r}^k}{r-1}\end{array}$ $\begin{array}{c}=\frac{a{r}^k-a+a{r}^{k+1}-a{r}^k}{r-1}\\ =\frac{a{r}^{k+1}-a}{r-1}\\ =\frac{a(r^{k+1}-1)}{r-1}\end{array}$ Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Q13. Prove the following by using the principle of mathematical induction for all n ∈ N: $(1+\frac{3}{1})(1+\frac{5}{4})(1+\frac{7}{9})\dots (1+\frac{(2n+1)}{n^2})=(n+1{)}^2$

Answer. Let the given statement be P(n), i.e., $P\left(n\right):(1+\frac{3}{1})(1+\frac{5}{4})(1+\frac{7}{9})\dots (1+\frac{(2n+1)}{n^2})=(n+1{)}^2$ For n = 1, we have $P\left(1\right):(1+\frac{3}{1})=4=(1+1{)}^2=2^2=4,$ Let P(k) be true for some positive integer k, i.e., $(1+\frac{3}{1})(1+\frac{5}{4})(1+\frac{7}{9})\dots (1+\frac{(2k+1)}{k^2})=(k+1{)}^2$ ……(1) We shall now prove that P(k + 1) is true. Consider $[(1+\frac{3}{1})(1+\frac{5}{4})(1+\frac{7}{9})\dots (1+\frac{(2k+1)}{k^2})]\{1+\frac{\left\{2\right(k+1)+1\}}{(k+1{)}^2}\}$ $=(k+1{)}^2(1+\frac{2(k+1)+1}{(k+1{)}^2})$ [ (Using (1) ] $\begin{array}{c}=(k+1{)}^2\left[\frac{(k+1{)}^2+2(k+1)+1}{(k+1{)}^2}\right]\\ =(k+1{)}^2+2(k+1)+1\\ =\left\{\right(k+1)+1{\}}^2\end{array}$ Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Q14. Prove the following by using the principle of mathematical induction for all n ∈ N: $(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3})\dots (1+\frac{1}{n})=(n+1)$

Answer. Let the given statement be P(n), i.e., P(n): $(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3})\dots (1+\frac{1}{n})=(n+1)$ For n = 1, we have $P\left(1\right):(1+\frac{1}{1})=2=(1+1)$, which is true Let P(k) be true for some positive integer k, i.e., $P\left(k\right):(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3})\dots (1+\frac{1}{k})=(k+1)$ ……….(1) We shall now prove that P(k + 1) is true. Consider $[(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3})\dots (1+\frac{1}{k})](1+\frac{1}{k+1})$ $=(k+1)(1+\frac{1}{k+1})$ [ Using(1) ] $\begin{array}{c}=(k+1)\left(\frac{(k+1)+1}{(k+1)}\right)\\ =(k+1)+1\end{array}$ Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Q15. Prove the following by using the principle of mathematical induction for all n ∈ N: $1^2+3^2+5^2+\dots +(2n-1{)}^2=\frac{n(2n-1)(2n+1)}{3}$

Answer. Let the given statement be P(n), i.e., $\begin{array}{c}P\left(n\right)=1^2+3^2+5^2+\dots +(2n-1{)}^2=\frac{n(2n-1)(2n+1)}{3}\\ \text{For}n=1,\text{we have}\\ P\left(1\right)=1^2=1=\frac{1(2.1-1)(2.1+1)}{3}=\frac{1.1.3}{3}=1,\text{which is true.}\end{array}$ Let P(k) be true for some positive integer k, i.e., $P\left(k\right)=1^2+3^2+5^2+\dots +(2k-1{)}^2=\frac{k(2k-1)(2k+1)}{3}$ …….(1) We shall now prove that P(k + 1) is true. Consider $\{1^2+3^2+5^2+\dots +(2k-1{)}^2\}+\left\{2\right(k+1)-1{\}}^2$ $=\frac{k(2k-1)(2k+1)}{3}+(2k+2-1{)}^2$ [ Using (1)] $\begin{array}{c}=\frac{k(2k-1)(2k+1)}{3}+(2k+1{)}^2\\ =\frac{k(2k-1)(2k+1)+3(2k+1{)}^2}{3}\\ =\frac{(2k+1)\left\{k\right(2k-1)+3(2k+1\left)\right\}}{3}\\ =\frac{(2k+1)\{2k^2-k+6k+3\}}{3}\end{array}$ $\begin{array}{c}=\frac{(2k+1)\{2k^2+5k+3\}}{3}\\ =\frac{(2k+1)\{2k^2+2k+3k+3\}}{3}\\ =\frac{(2k+1)\left\{2k\right(k+1)+3(k+1\left)\right\}}{3}\end{array}$ $\begin{array}{c}=\frac{(2k+1)(k+1)(2k+3)}{3}\\ =\frac{(k+1)\left\{2\right(k+1)-1\}\left\{2\right(k+1)+1\}}{3}\end{array}$ Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Q16. Prove the following by using the principle of mathematical induction for all n ∈ N: $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\dots +\frac{1}{(3n-2)(3n+1)}=\frac{n}{(3n+1)}$

Answer. Let the given statement be P(n), i.e., $\begin{array}{c}P\left(n\right):\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\dots +\frac{1}{(3n-2)(3n+1)}=\frac{n}{(3n+1)}\\ \text{For}n=1,\text{we have}\\ P\left(1\right)=\frac{1}{1.4}=\frac{1}{3.1+1}=\frac{1}{4}=\frac{1}{1.4},\text{which is true.}\end{array}$ Let P(k) be true for some positive integer k, i.e., $P\left(k\right)=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\dots +\frac{1}{(3k-2)(3k+1)}=\frac{k}{3k+1}$ ………..(1) We shall now prove that P(k + 1) is true. Consider $\{\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\dots +\frac{1}{(3k-2)(3k+1)}\}+\frac{1}{\left\{3\right(k+1)-2\}\left\{3\right(k+1)+1\}}$ $=\frac{k}{3k+1}+\frac{1}{(3k+1)(3k+4)}$ [ Using (1)] $\begin{array}{c}=\frac{1}{(3k+1)}\{k+\frac{1}{(3k+4)}\}\\ =\frac{1}{(3k+1)}\left\{\frac{k(3k+4)+1}{(3k+4)}\right\}\\ =\frac{1}{(3k+1)}\left\{\frac{3k^2+4k+1}{(3k+4)}\right\}\end{array}$ $\begin{array}{c}=\frac{1}{(3k+1)}\left\{\frac{3k^2+3k+k+1}{(3k+4)}\right\}\\ =\frac{(3k+1)(k+1)}{(3k+1)(3k+4)}\\ =\frac{(k+1)}{3(k+1)+1}\end{array}$ Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Q17. Prove the following by using the principle of mathematical induction for all n ∈ N: $\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\dots +\frac{1}{(2n+1)(2n+3)}=\frac{n}{3(2n+3)}$

Answer. Let the given statement be P(n), i.e., $P\left(n\right):\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\dots +\frac{1}{(2n+1)(2n+3)}=\frac{n}{3(2n+3)}$ For n = 1, we have $P\left(1\right):\frac{1}{3.5}=\frac{1}{3(2.1+3)}=\frac{1}{3.5}$ , which is true. Let P(k) be true for some positive integer k, i.e., $P\left(k\right):\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\dots +\frac{1}{(2k+1)(2k+3)}=\frac{k}{3(2k+3)}$ ……(1) We shall now prove that P(k + 1) is true. Consider $[\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\dots +\frac{1}{(2k+1)(2k+3)}]+\frac{1}{\left\{2\right(k+1)+1\}\left\{2\right(k+1)+3\}}$ $=\frac{k}{3(2k+3)}+\frac{1}{(2k+3)(2k+5)}$ $\left[U\mathrm{sing}\right(1\left)\right]$ $\begin{array}{c}=\frac{1}{(2k+3)}[\frac{k}{3}+\frac{1}{(2k+5)}]\\ =\frac{1}{(2k+3)}\left[\frac{k(2k+5)+3}{3(2k+5)}\right]\\ =\frac{1}{(2k+3)}\left[\frac{2k^2+5k+3}{3(2k+5)}\right]\end{array}$ $\begin{array}{c}=\frac{1}{(2k+3)}\left[\frac{2k^2+2k+3k+3}{3(2k+5)}\right]\\ =\frac{1}{(2k+3)}\left[\frac{2k(k+1)+3(k+1)}{3(2k+5)}\right]\\ =\frac{(k+1)(2k+3)}{3(2k+3)(2k+5)}\\ =\frac{(k+1)}{3\left\{2\right(k+1)+3\}}\end{array}$ Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Q18. Prove the following by using the principle of mathematical induction for all n ∈ N: $1+2+3+\dots +n<\frac{1}{8}(2n+1{)}^2$

Answer. Let the given statement be P(n), i.e., $P\left(n\right):1+2+3+\dots +n<\frac{1}{8}(2n+1{)}^2$ It can be noted that P(n) is true for n = 1 since $1<\frac{1}{8}(2.1+1{)}^2=\frac{9}{8}$ Let P(k) be true for some positive integer k, i.e., $1+2+\dots +k<\frac{1}{8}(2k+1{)}^2$ ……(1) We shall now prove that P(k + 1) is true whenever P(k) is true. Consider $(1+2+\dots +k)+(k+1)<\frac{1}{g}(2k+1{)}^2+(k+1)$ $\left[\text{Using}\right(1\left)\right]$ $\begin{array}{c}<\frac{1}{8}\left\{\right(2k+1{)}^2+8(k+1)\}\\ <\frac{1}{8}\{4k^2+4k+1+8k+8\}\\ <\frac{1}{8}\{4k^2+12k+9\}\\ <\frac{1}{8}(2k+3{)}^2\\ <\frac{1}{8}\left\{2\right(k+1)+1{\}}^2\end{array}$ Hence, $(1+2+3+\dots +k)+(k+1)<\frac{1}{8}(2k+1{)}^2+(k+1)$ Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Q19. Prove the following by using the principle of mathematical induction for all n ∈ N: $n(n+1)(n+5)\text{is a multiple of}3$

Answer. Let the given statement be P(n), i.e., P(n): n (n + 1) (n + 5), which is a multiple of 3. It can be noted that P(n) is true for n = 1 since 1 (1 + 1) (1 + 5) = 12, which is a multiple of 3. Let P(k) be true for some positive integer k, i.e., k (k + 1) (k + 5) is a multiple of 3. ∴k (k + 1) (k + 5) = 3m, where m ∈ N … (1) We shall now prove that P(k + 1) is true whenever P(k) is true. Consider $\begin{array}{c}(k+1)\left\{\right(k+1)+1\}\left\{\right(k+1)+5\}\\ =(k+1)(k+2)\left\{\right(k+5)+1\}\\ =(k+1)(k+2)(k+5)+(k+1)(k+2)\\ =\left\{k\right(k+1\left)\right(k+5)+2(k+1\left)\right(k+5\left)\right\}\\ =3m+(k+1)\left\{2\right(k+5)+(k+2\left)\right\}\\ =3m+(k+1)\{2k+10+k+2\}\\ =3m+(k+1)\{2k+12)\\ =3m+3(k+1)(k+4)\end{array}$ $\begin{array}{c}=3\{m+(k+1\left)\right(k+4\left)\right\}=3\times q,\text{where}q=\{m+(k+1\left)\right(k+4\left)\right\}\text{is some natural number}\\ \text{Therefore,}(k+1)\left\{\right(k+1)+1\}\left\{\right(k+1)+5\}\text{is a multiple of}3.\end{array}$ Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Q20. Prove the following by using the principle of mathematical induction for all n ∈ N: ${10}^{1n-1}+1\text{is divisible by}11$

Answer. Let the given statement be P(n), i.e., $\begin{array}{c}P\left(n\right):{10}^{2n-1}+1\text{is divisible by}11.\\ \text{It can be observed that}P\left(n\right)\text{is true for}n=1\text{since}P\left(1\right)={10}^{2.1-1}+1=\\ 11,\text{which is divisible by}11.\end{array}$ Let P(k) be true for some positive integer k, i.e., $\begin{array}{c}{10}^{2k-1}+1\text{is divisible by}11.\\ \therefore {10}^{2k-1}+1=11m,\text{where}m\in N\dots \left(1\right)\end{array}$ We shall now prove that P(k + 1) is true whenever P(k) is true. Consider $\begin{array}{c}{10}^{2(k+1)-1}+1\\ ={10}^{2k+2-1}+1\\ ={10}^{2k+1}+1\\ ={10}^2({10}^{2k-1}+1-1)+1\\ ={10}^2({10}^{2k-1}+1)-{10}^2+1\end{array}$ $={10}^2\cdot 11m-100+1$ $\left[\text{Using}\right(1\left)\right]$ Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Q21. Prove the following by using the principle of mathematical induction for all n ∈ N: $x^{2x}-y^{3n}\text{is divisible by}x+y$

Answer. Let the given statement be P(n), i.e., $\begin{array}{c}P\left(n\right):x^{2n}-y^{2n}\text{is divisible by}x+y\text{.}\\ \text{It can be observed that}P\left(n\right)\text{is true for}n=1\text{.}\\ \text{This is so because}{x}^2\times 1-y^2\times 1=x^2-y^2=(x+y)(x-y)\text{is divisible by}\\ (x+y).\end{array}$ Let P(k) be true for some positive integer k, i.e., $\begin{array}{c}{x}^{2k}-y^{2k}\text{is divisible by}x+y.\\ \therefore x^{2k}-y^{2k}=m(x+y),\text{where}m\in N\dots \left(1\right)\end{array}$ We shall now prove that P(k + 1) is true whenever P(k) is true. Consider $\begin{array}{c}{x}^{2(k+1)}-y^{2(k+1)}\\ =x^{2k}\cdot x^2-y^{2k}\cdot y^2\\ =x^2(x^{2k}-y^{2k}+y^{2k})-y^{2k}\cdot y^2\end{array}$ $=x^2\left\{m\right(x+y)+y^{2k}\}-y^{2k}\cdot y^2\left[\text{Using}\right(1\left)\right]$ $\begin{array}{c}=m(x+y)x^2+y^{2k}\cdot x^2-y^{2k}\cdot y^2\\ =m(x+y)x^2+y^{2k}(x^2-y^2)\\ =m(x+y)x^2+y^{2k}(x+y)(x-y)\\ =(x+y)\{m{x}^2+y^{2k}(x-y\left)\right\},\text{which is a factor of}(x+y)\end{array}$ Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Q22. Prove the following by using the principle of mathematical induction for all n ∈ N: $3^{2n+2}-8n-9\text{is divisible by}8$

Answer. Let the given statement be P(n), i.e., $P\left(n\right):3^{2n+2}-8n-9\text{is divisible by}8$ $\begin{array}{c}\text{It can be observed that}P\left(n\right)\text{is true for}n=1\text{since}{3}^{2\times 1+2}-8\times 1-9=\\ 64,\text{which is divisible by}8\text{.}\end{array}$ Let P(k) be true for some positive integer k, i.e., $\begin{array}{c}{3}^{2k+2}-8k-9\text{is divisible by}8\text{.}\\ \therefore 3^{2k+2}-8k-9=8m;\text{where}m\in N\dots \left(1\right)\end{array}$ We shall now prove that P(k + 1) is true whenever P(k) is true. Consider $\begin{array}{c}{3}^{2(k+1)+2}-8(k+1)-9\\ =3^{2k+2}\cdot 3^2-8k-8-9\\ =3^2(3^{2k+2}-8k-9+8k+9)-8k-17\\ =3^2(3^{2k+2}-8k-9)+3^2(8k+9)-8k-17\\ =9.8m+9(8k+9)-8k-17\end{array}$ $\begin{array}{c}=9.8m+9(8k+9)-8k-17\\ =9.8m+72k+81-8k-17\\ =9.8m+64k+64\\ =8(9m+8k+8)\\ =8r,\text{where}r=(9m+8k+8)\text{is a natural number}\\ \text{Therefore,}{3}^{2(k+1)+2}-8(k+1)-9\text{is divisible by}8\text{.}\end{array}$ Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Q23. Prove the following by using the principle of mathematical induction for all n ∈ N: ${41}^n-{14}^n\text{is a multiple of}27$

Answer. Let the given statement be P(n), i.e., $P\left(n\right):{41}^n-{14}^n\text{is a multiple of}27$ It can be observed that P(n) is true for n = 1 since ${41}^1-{14}^1=27$ which is a multiple of 27. Let P(k) be true for some positive integer k, i.e., $\begin{array}{c}{41}^k-{14}^k\text{is a multiple of}27\\ \therefore {41}^k-{14}^k=27m,\text{where}m\in N\dots \text{(1)}\end{array}$ We shall now prove that P(k + 1) is true whenever P(k) is true. Consider $\begin{array}{c}{41}^{k+1}-{14}^{k+1}\\ ={41}^k\cdot 41-{14}^k\cdot 14\\ =41({41}^k-{14}^k+{14}^k)-{14}^k\cdot 14\\ =41({41}^k-{14}^k)+{41.14}^k-{14}^k\cdot 14\\ =41.27m+{14}^k(41-14)\end{array}$ $\begin{array}{c}=41.27m+{27.14}^k\\ =27(41m-{14}^k)\end{array}$ $\begin{array}{c}=27\times r,\text{where}r=(41m-{14}^k)\text{is a natural number}\\ \text{Therefore,}{41}^{k+1}-{14}^{k+1}\text{is a multiple of}27\end{array}$ Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Q24. Prove the following by using the principle of mathematical induction for all n ∈ N: $(2n+7)<(n+3{)}^2$

Answer. Let the given statement be P(n), i.e., $P\left(n\right):(2n+7)<(n+3{)}^2$ $\begin{array}{c}\text{It can be observed that}P\left(n\right)\text{is true for}n=1\text{since}2.1+7=9<(1+3{)}^2\\ =16,\text{which is true.}\end{array}$ Let P(k) be true for some positive integer k, i.e., $(2k+7)<(k+3{)}^2\dots (1)$ We shall now prove that P(k + 1) is true whenever P(k) is true. Consider $\left\{2\right(k+1)+7\}=(2k+7)+2$ $\therefore \left\{2\right(k+1)+7\}=(2k+7)+2<(k+3{)}^2+2[\text{using}\left(1\right)]$ $\begin{array}{c}2(k+1)+7<k^2+6k+9+2\\ 2(k+1)+7<k^2+6k+11\\ \text{Now,}{k}^2+6k+11<k^2+8k+16\\ \therefore 2(k+1)+7<(k+4{)}^2\\ 2(k+1)+7<\left\{\right(k+1)+3{\}}^2\end{array}$ Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.