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NCERT Solutions Class 11 Maths Chapter 4 Principle of Mathematical Induction – Here are all the NCERT solutions for Class 11 Maths Chapter 4. This solution contains questions, answers, images, explanations of the complete chapter 4 titled Of Principle of Mathematical Induction taught in Class 11. If you are a student of Class 11 who is using NCERT Textbook to study Maths, then you must come across chapter 4 Principle of Mathematical Induction After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction in one place.

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NCERT Solutions Class 11 Maths Chapter 4 Principle of Mathematical Induction

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Class 11
Subject Maths
Book Mathematics
Chapter Number 4
Chapter Name

Principle of Mathematical Induction

NCERT Solutions Class 11 Maths chapter 4 Principle of Mathematical Induction

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Question & Answer

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

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

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

Ans : \(\begin{array}{l}{\text { Let the given statement be } P(n), \text { i.e., }} \\ {\text { P(n): } 1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+\ldots n}=\frac{2 n}{n+1}} \\ {\text { For } n=1, \text { we have }} \\ {P(1) : 1=\frac{2.1}{1+1}=\frac{2}{2}=1} \\ {\text { Let } P(k) \text { be true for some positive integer } k, \text { i.e., }}\end{array}\)
\(1+\frac{1}{1+2}+\ldots+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+\ldots+k}=\frac{2 k}{k+1}\)............(i)
\(\begin{array}{l}{\text { We shall now prove that } \mathrm{P}(k+1) \text { is true. }} \\ {\text { Consider }} \\ {1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+\ldots+k}+\frac{1}{1+2+2+k+(k+1)}}\end{array}\)
\(\begin{array}{l}{=\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+. k}\right)+\frac{1}{1+2+3+\ldots+k+(k+1)}} \\ {=\frac{2 k}{k+1}+\frac{1}{1+2+3+\ldots+k+(k+1)} \quad \quad[\text { Using (i) }]}\end{array}\)
\(=\frac{2 k}{k+1}+\frac{1}{\left(\frac{(k+1)(k+1+1)}{2}\right)}\)     \(\left[1+2+3+\ldots+n=\frac{n(n+1)}{2}\right]\)
\(\begin{array}{l}{=\frac{2 k}{(k+1)}+\frac{2}{(k+1)(k+2)}} \\ {=\frac{2}{(k+1)}\left(k+\frac{1}{k+2}\right)}\end{array}\) 
Q.4: Prove the following by using the principle of mathematical induction for all n ∈ N:
\(1.2.3+2.3 .4+\ldots+n(n+1)(n+2)=\frac{n(n+1)(n+2)(n+3)}{4}\)

Ans : \(\begin{array}{l}{\text { Let the given statement be } P(n), \text { i.e., }} \\ {P(n) : 1.2 .3+2.3 .4+\ldots+n(n+1)(n+2)=\frac{n(n+1)(n+2)(n+3)}{4}}\end{array}\)
\(\begin{array}{l}{\text { For } n=1, \text { we have }} \\ {P(1) : 1.2 \cdot 3=6=\frac{1(1+1)(1+2)(1+3)}{4}=\frac{1.23 .4}{4}=6}\end{array}\) , which is true
Let P(k) be true for some positive integer k, i.e., 
\(1.2 .3+2.3 .4+\ldots+k(k+1)(k+2)=\frac{k(k+1)(k+2)(k+3)}{4}\)    ……………(i)
\(\begin{array}{l}{\text { We shall now prove that } P(k+1) \text { is true. }} \\ {\text { Consider }} \\ {1.2 .3+2.3 .4+\ldots+k(k+1)(k+2)+(k+1)(k+2)(k+3)} \\ {=\{1.2 .3+2.3 .4+\ldots+k(k+1)(k+2)\}+(k+1)(k+2)(k+3)}\end{array}\)
\(\begin{array}{l}{=\frac{k(k+1)(k+2)(k+3)}{4}+(k+1)(k+2)(k+3) \quad[\text { Using }(i)]} \\ {=(k+1)(k+2)(k+3)\left(\frac{k}{4}+1\right)} \\ {=\frac{(k+1)(k+2)(k+3)(k+4)}{4}} \\ {=\frac{(k+1)(k+1+1)(k+1+2)(k+1+3)}{4}}\end{array}\)
\(\begin{array}{l}{\text { Thus, } P(k+1) \text { is true whenever } P(k) \text { is true. }} \\ {\text { Hence, by the principle of mathematical induction, statement } P(n) \text { is true for all natural }} \\ {\text { numbers l.e., } n \text { . }}\end{array}\) 
Q.5: Prove the following by using the principle of mathematical induction for all n ∈ N:
\(1.3+2.3^{2}+3.3^{3}+\ldots+n 3^{n}=\frac{(2 n-1) 3^{n+1}+3}{4}\)

Ans : \(\begin{array}{l}{\text { Let the given statement be } P(n), \text { i.e., }} \\ {1.3+2.3^{2}+3.3^{\circ}+\ldots+n 3^{n}=\frac{(2 n-1) 3^{n+1}+3}{4}}\end{array}\)
\(\begin{array}{l}{P(n) :} \\ {\text { For } n=1, \text { we have }}\end{array}\)
\(P(1) : 1.3=3 \quad=\frac{(2.1-1) 3^{1+1}+3}{4}=\frac{3^{2}+3}{4}=\frac{12}{4}=3\)
\(\begin{array}{l}{\text { Let } P(k) \text { be true for some positive integer } k, \text { i.e., }} \\ {1.3+2.3^{2}+3.3^{3}+\ldots+k 3^{4}=\frac{(2 k-1) 3^{k+1}+3}{4}} \\ {\text { We shall now prove that } P(k+1) \text { is true. }}\end{array}\)  ……….(i)
Consider 
\(\begin{array}{l}{1.3+2.3^{2}+3.3^{3}+\ldots+k 3^{k}+(k+1) 3^{k+1}} \\ {=\left(1.3+2.3^{2}+3.3^{3}+\ldots+k .3^{k}\right)+(k+1) 3^{k+1}}\end{array}\)
\(=\frac{(2 k-1) 3^{k+1}+3}{4}+(k+1) 3^{k+1} \quad[\text { Using }(\mathbf{i})]\)
\(\begin{array}{l}{=\frac{(2 k-1) 3^{3+1}+3+4(k+1) 3^{k+1}}{4}} \\ {=\frac{3^{t+1}\{2 k-1+4(k+1)\}+3}{4}}\end{array}\)
\(\begin{aligned} &=\frac{3^{k+1}\{6 k+3\}+3}{4} \\ &=\frac{3^{k+1} \cdot 3\{2 k+1\}+3}{4} \end{aligned}\)
\(\begin{array}{l}{=\frac{3^{(k+1)+1}\{2 k+1\}+3}{4}} \\ {=\frac{\{2(k+1)-1\} 3^{(k+1)+1}+3}{4}}\end{array}\)
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.

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