Principle of Mathematical Induction Class 11 NCERT Solutions Chapter 4

Principle of Mathematical Induction Class 11 NCERT Solutions provides clear and detailed answers to all the questions of Chapter 4. This chapter introduces the method of mathematical induction, which is used to prove statements involving natural numbers. Here, you will find step-by-step solutions that explain the base step and inductive step clearly, helping you understand the logic behind proofs and score better in your exams.

Principle of Mathematical Induction Class 11 NCERT

Exercise 4.1 – Principle of Mathematical Induction Class 11 NCERT

We prove both results using Mathematical Induction.

1. Prove that

1+3+32+⋯+3n−1=3n−121 + 3 + 3^2 + \dots + 3^{n-1} = \frac{3^n – 1}{2}

LHS = 11

RHS =31−12=3−12=1\frac{3^1 – 1}{2} = \frac{3 – 1}{2} = 1

LHS = RHS ✔
True for n=1n = 1

Step 2: Induction Hypothesis

Assume true for n=kn = k1+3+32+⋯+3k−1=3k−121 + 3 + 3^2 + \dots + 3^{k-1} = \frac{3^k – 1}{2}

Step 3: Prove for n=k+1n = k+1

Consider LHS for k+1k+11+3+32+⋯+3k−1+3k1 + 3 + 3^2 + \dots + 3^{k-1} + 3^k

Using induction hypothesis:=3k−12+3k= \frac{3^k – 1}{2} + 3^k

Take LCM:=3k−1+2⋅3k2= \frac{3^k – 1 + 2 \cdot 3^k}{2}=3⋅3k−12= \frac{3 \cdot 3^k – 1}{2}=3k+1−12= \frac{3^{k+1} – 1}{2}

Which is RHS for k+1k+1

✔ Hence true for k+1k+1

Therefore, by mathematical induction,

1+3+32+⋯+3n−1=3n−121 + 3 + 3^2 + \dots + 3^{n-1} = \frac{3^n – 1}{2}

2. Prove that

13+23+33+⋯+n3=(n(n+1)2)21^3 + 2^3 + 3^3 + \dots + n^3 = \left(\frac{n(n+1)}{2}\right)^2

Step 1: Base Case (n = 1)

LHS = 13=11^3 = 113=1

RHS =(1(1+1)2)2=(22)2=1\left(\frac{1(1+1)}{2}\right)^2 = \left(\frac{2}{2}\right)^2 = 1(21(1+1)​)2=(22​)2=1

LHS = RHS ✔

Step 2: Induction Hypothesis

Assume true for n=kn = kn=k:13+23+⋯+k3=(k(k+1)2)21^3 + 2^3 + \dots + k^3 = \left(\frac{k(k+1)}{2}\right)^2

Step 3: Prove for n=k+1n = k+1

Consider LHS for k+1k+113+23+⋯+k3+(k+1)31^3 + 2^3 + \dots + k^3 + (k+1)^3

Using hypothesis:=(k(k+1)2)2+(k+1)3= \left(\frac{k(k+1)}{2}\right)^2 + (k+1)^3=k2(k+1)24+(k+1)3= \frac{k^2 (k+1)^2}{4} + (k+1)^3

Take common factor (k+1)2(k+1)^2=(k+1)2(k24+(k+1))= (k+1)^2 \left( \frac{k^2}{4} + (k+1) \right)=(k+1)2(k2+4k+44)= (k+1)^2 \left( \frac{k^2 + 4k + 4}{4} \right)=(k+1)2((k+2)24)= (k+1)^2 \left( \frac{(k+2)^2}{4} \right)=(k+1)2(k+2)24= \frac{(k+1)^2 (k+2)^2}{4}=((k+1)(k+2)2)2= \left(\frac{(k+1)(k+2)}{2}\right)^2

Which is RHS for k+1k+1

✔ Hence true for k+1k+1

Therefore, by mathematical induction,

13+23+33+⋯+n3=(n(n+1)2)21^3 + 2^3 + 3^3 + \dots + n^3 = \left(\frac{n(n+1)}{2}\right)^2

Prove the following by using the principle of mathematical induction for all n∈Nn \in \mathbb{N}.

1.

1+3+32+⋯+3n−1=3n−121 + 3 + 3^2 + \dots + 3^{n-1} = \frac{3^n – 1}{2}

2.

13+23+33+⋯+n3=(n(n+1)2)21^3 + 2^3 + 3^3 + \dots + n^3 = \left( \frac{n(n+1)}{2} \right)^2

3.

1+11+2+11+2+3+⋯+11+2+3+⋯+n=2nn+11 + \frac{1}{1+2} + \frac{1}{1+2+3} + \dots + \frac{1}{1+2+3+\dots+n} = \frac{2n}{n+1}

4.

1⋅2⋅3+2⋅3⋅4+⋯+n(n+1)(n+2)=n(n+1)(n+2)(n+3)41\cdot2\cdot3 + 2\cdot3\cdot4 + \dots + n(n+1)(n+2) = \frac{n(n+1)(n+2)(n+3)}{4}

5.

1⋅3+2⋅32+3⋅33+⋯+n⋅3n=(2n−1)3n+1+341\cdot3 + 2\cdot3^2 + 3\cdot3^3 + \dots + n\cdot3^n = \frac{(2n-1)3^{n+1} + 3}{4}

6.

1⋅2+2⋅3+3⋅4+⋯+n(n+1)=n(n+1)(n+2)31\cdot2 + 2\cdot3 + 3\cdot4 + \dots + n(n+1) = \frac{n(n+1)(n+2)}{3}

7.

1⋅3+3⋅5+5⋅7+⋯+(2n−1)(2n+1)=n(4n2+6n−1)31\cdot3 + 3\cdot5 + 5\cdot7 + \dots + (2n-1)(2n+1) = \frac{n(4n^2 + 6n – 1)}{3}

8.

1⋅2+2⋅22+3⋅23+⋯+n⋅2n=(n−1)2 n+1+21\cdot2 + 2\cdot2^2 + 3\cdot2^3 + \dots + n\cdot2^n = (n-1)2^{\,n+1} + 2

9.

12+14+18+⋯+12n=1−12n\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^n} = 1 – \frac{1}{2^n}

10.

12⋅5+15⋅8+18⋅11+⋯+1(3n−1)(3n+2)=n(6n+4)\frac{1}{2\cdot5} + \frac{1}{5\cdot8} + \frac{1}{8\cdot11} + \dots + \frac{1}{(3n-1)(3n+2)} = \frac{n}{(6n+4)}

11.

11⋅2⋅3+12⋅3⋅4+13⋅4⋅5+⋯+1n(n+1)(n+2)=n(n+3)4(n+1)(n+2)\frac{1}{1\cdot2\cdot3} + \frac{1}{2\cdot3\cdot4} + \frac{1}{3\cdot4\cdot5} + \dots + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}

12.

a+ar+ar2+⋯+ar n−1=a(rn−1)r−1a + ar + ar^2 + \dots + ar^{\,n-1} = \frac{a(r^n – 1)}{r – 1}

13.

(1+31)(1+54)(1+79)⋯(1+2n+1n2)=(n+1)2\left(1 + \frac{3}{1}\right) \left(1 + \frac{5}{4}\right) \left(1 + \frac{7}{9}\right) \cdots \left(1 + \frac{2n+1}{n^2}\right) = (n+1)^2

14.

(1+11)(1+12)(1+13)⋯(1+1n)=n+1\left(1 + \frac{1}{1}\right) \left(1 + \frac{1}{2}\right) \left(1 + \frac{1}{3}\right) \cdots \left(1 + \frac{1}{n}\right) = n+1

15.

12+32+52+⋯+(2n−1)2=n(2n−1)(2n+1)31^2 + 3^2 + 5^2 + \dots + (2n-1)^2 = \frac{n(2n-1)(2n+1)}{3}

16.

11⋅4+14⋅7+17⋅10+⋯+1(3n−2)(3n+1)=n3n+1\frac{1}{1\cdot4} + \frac{1}{4\cdot7} + \frac{1}{7\cdot10} + \dots + \frac{1}{(3n-2)(3n+1)} = \frac{n}{3n+1}

17.

13⋅5+15⋅7+17⋅9+⋯+1(2n+1)(2n+3)=n3(2n+3)\frac{1}{3\cdot5} + \frac{1}{5\cdot7} + \frac{1}{7\cdot9} + \dots + \frac{1}{(2n+1)(2n+3)} = \frac{n}{3(2n+3)}

18.

1+2+3+⋯+n<18(2n+1)21 + 2 + 3 + \dots + n < \frac{1}{8}(2n+1)^2

19.

n(n+1)(n+5) is a multiple of 3.n(n+1)(n+5) \text{ is a multiple of } 3.

20.

102n+1+1 is divisible by 11.10^{2n+1} + 1 \text{ is divisible by } 11.

21.

x2n−y2n is divisible by x+y.x^{2n} – y^{2n} \text{ is divisible by } x + y.

22.

32n+2−8n−9 is divisible by 8.3^{2n+2} – 8n – 9 \text{ is divisible by } 8.

23.

4 n+1−14n is a multiple of 27.4^{\,n+1} – 14^n \text{ is a multiple of } 27.

24.

(2n+7)<(n+3)2.(2n+7) < (n+3)^2.

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