## NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction Exercise 4.1

Chapter 4 Principle of Mathematical Induction Exercise 4.1 NCERT Solutions for Class 11 Maths will help you in finding solutions to difficult questions without any hassle. The Class 11 Maths NCERT Solutions prepared by us will help you in building the basics of the chapter and understanding the practical use of formulas.

1. Prove the following by using the principle of mathematical induction for all nâˆˆÂ N1 + 3 + 3

^{2}Â + â€¦â€¦â€¦â€¦. + 3

^{n-1}Â = (3

^{n}Â â€“ 1)/2

**Answer**

Let P(n) be the given statement.

i.e., P(n) : 1 + 3 + 3

^{2}Â + â€¦â€¦â€¦â€¦â€¦.. + 3

^{n-1}Â = (3

^{n}Â â€“ 1)/2

Putting n = 1, P(1) = ( 3 â€“ 1)/2 = 1

P(n) is true for n = 1

Assume that P(k) is true.

i.e., P(k): 1 + 3 + 3

^{2}Â + â€¦â€¦â€¦.. + 3

^{k -1}

= (3

^{kÂ }Â â€“ 1)/2 â€¦.(i)

We shall prove that P(k + 1) is true whenever P(k) is true.

P(k + 1) : 1 + 3 + 3

^{2}Â + â€¦â€¦â€¦â€¦ + 3

^{(k + 1) - 1}

= (3

^{k + 1}Â â€“ 1)/2

[1 + 3 + 3

^{2}Â + â€¦â€¦.. + 3

^{k-1}] + 3

^{k}Â = (3

^{k}Â â€“ 1)/2 + 3

^{k}

[using (i)]

(3

^{k}Â â€“ 1 + 2.3

^{k})/2 = ((1 + 2)3

^{k}Â â€“ 1)/2 = (3.3

^{k}Â -1)/2

= (3

^{k+1}Â â€“ 1)/2

P(k + 1) is also true whenever P(k) is true.

Hence, P(n) is true for all nÂ âˆˆÂ N

2. 1

^{3}Â + 2

^{3}Â + 3

^{3}Â + â€¦â€¦â€¦. + n

^{3}Â = ((n(n + 1))/2)

^{2}

**Answer**

Let P(n) : 1

^{3}Â + 2

^{3}Â + 3

^{3}Â + â€¦â€¦â€¦. + n

^{3}

= (n

^{2}(n + 1)

^{2})/4 â€¦â€¦..(i)

For n = 1, L.H.S = 1

^{3}Â = 1

and RHS = (1

^{2}Â (1+1)

^{2})/4 = 1.4/4 = 1

LHS = RHS i.e., P(1) is true

Let us suppose that P(k) is true.

Putting n = k in (i) we have

P(k) : 1

^{3}Â + 2

^{3}Â + 3

^{3}Â + â€¦â€¦â€¦â€¦ + k

^{3}Â = (k

^{2}(k + 1)

^{2})/4 â€¦..(ii)

We have to prove that P(k + 1) is true whenever

P(k) is true.

P(k + 1) : 1

^{3}Â + 2

^{3}+ 3

^{3}Â + â€¦â€¦â€¦. + k

^{3}Â + (k + 1)

^{3}

= (k

^{2}Â (k + 1)

^{2})/4 + (k + 1)

^{3}[using (i)]

= (k + 1)

^{2}[(k

^{2}Â + 4(k + 1))/4]

= ((k + 1)

^{2}(k

^{2}Â + 4k + 4))/4

= ((k +1 )

^{2}(k + 2)

^{2})/4 =

P(n) is true for n = k + 1 i.e., P(k + 1) is true

By principle of mathematical induction.

P(n) is true for all natural numbers n

3. 1 + 1/(1 + 2) + 1/(1 + 2+ 3) + â€¦â€¦+ 1/( 1+2+3+.....+n) = 2n/n +1

**Answer**

Let P(n): 1 + 1/(1 + 2) + 1/(1 + 2+ 3) + â€¦â€¦.. + 1/(1 + 2+ 3+ â€¦â€¦ + n) = 2n/n + 1 â€¦. (i)

Putting n = 1,

LHS = 1 and RHS = (2.1)/(1+1) = 2/2 = 1

LHS = RHS âˆ´Â P(1) is true

Let P(k) be true.

Putting n = k then

P(k): 1 + 1/(1 + 2) + â€¦â€¦â€¦â€¦. + 1/(1 + 2+ 3 + â€¦â€¦. + k)

=2k/k + 1 â€¦â€¦..(ii)

Now we shall prove that P(k + 1) is true whenever P(k) is true.

P(k + 1):

1 + 1/(1+2) + â€¦â€¦â€¦. + 1/(1 + 2 + 3 + â€¦..+ k) + 1/(1 + 2 + 3 +â€¦â€¦..(k + 1))

= 2k/(k + ) + 1/(1 + 2+ 3 + â€¦(k + 1) [from (ii)]

= 2k/(k + 1) + 1/((k + 1)(k + 2))/2

=(2k(k + 2) + 2)/((k + 1)(k + 2)) = (2[k(k + 2) + 1])/(k + 1)(k + 2))

=

P(k + 1) is true whenever P(k) is true.

Hence, P(k) is true for all nÂ âˆˆÂ N

4.Â 1.2.3 + 2.3.4 + â€¦â€¦â€¦.. + n(n + 1)(n + 2) =(n(n + 1)(n + 2)(n + 3))/4

**Answer**

Let P(n) : 1.2.3 + 2.3.4 + â€¦â€¦â€¦â€¦ + n(n+ 1)(n + 2) = (n(n +1)(n + 2)(n + 3))/4Â Â Â Â Â â€¦â€¦â€¦.(i)

For n = 1, LHS = 1.2.3 = 6 and

RHS = (1(1 +1)(1 + 2)(1 + 3))/4 = (1 Ã— 2 Ã— 3 Ã— 4)/4 = 6

LHS = RHS i.e., P(1) is true.

Let P(k) is true.

P(k): 1.2.3 + 2.3.4 + â€¦â€¦..+ k(k + 1)(k + 2) = (k(k + 1)(k + 2)(k + 3)/4Â Â â€¦â€¦â€¦..(ii)

We have to prove that

P(k + 1) is true whenever P(k) is true.

P(k + 1): [1.2.3 + 2.3.4 + 3.4.5 + â€¦â€¦â€¦â€¦. + k(k + 1)(k + 2)]+(k + 1)(k + 2)(k + 3)

= (k(k + 1)(k + 2)(k + 3)/4Â + (k + 1)(k + 2)(k + 3)

= (k + 1)(k + 2)(k + 3)(k/4 + 1) [from(ii)]

= ((k + 1)(k + 2)(k + 3)(k + 4))/4Â Â Â Â Â â€¦â€¦(iii)

P(n) is true for n = k + 1 i.e., P(k + 1) is true whenever P(k) is true.

By principle of mathematical induction,

P(n) is true for all natural numbers n.

5. 1.3 + 2.3

^{2}Â + 3.3

^{3}Â + â€¦â€¦â€¦+ n.3

^{n}Â = ((2n -1)3

^{n+1}Â + 3))/4

**Answer**

Let P(n): 1.3 + 2.3

^{2}Â + 3.3

^{3}Â + â€¦â€¦.+ n.3

^{n}Â = ((2n â€“ 1)3

^{n+1}Â + 3)/4

Putting n = 1, P(1): LHS = 1.3 = 3

RHS = ((2 -1)3

^{2}Â + 3)/4 = 12/4 = 3 = LHS

This shows P(n) is true for n = 1

Let P(n) be true for n = k

i.e., P(k): 1.3 + 2.3

^{2}Â + 3.3

^{3}Â + â€¦â€¦ + k.3

^{k}Â = ((2k -1)3

^{k+1}Â + 3)/4 â€¦â€¦..(i)

we have to prove that P(k + 1) is true whenever P(k) is true.

P(k + 1): 1.3 + 2.3

^{2}Â + â€¦.. +(k + 1)3

^{k+1}Â = [1.3 + 2.3

^{2}Â + 3.3

^{3}Â + â€¦â€¦. + k.3

^{k}] +(k + 1)3

^{k + 1}

= ((2k â€“ 1).3

^{k+1}Â + 3)/4 + (k + 1).3

^{k+1}[using (i)]

=((2k â€“ 1)3

^{k+1}Â + 3 + 4(k + 1)3

^{k+1})/4

= (3

^{k+1}Â [2k â€“ 1 + 4(k + 1)] + 3)/4

= (3

^{k+1}(6k + 3) + 3)/4 = ((2k + 1)3

^{k+2}Â + 3)/4

=([2(k + 1)- 1]3

^{(k+1)+1}Â + 3)/4

This shows P(n) is true for n = k + 1

i.e., P(k + 1) is true whenever P(k) is true.

Hence, P(n) is true for all value nÂ âˆˆÂ N

6.Â 1.2 + 2.3 + 3.4+ â€¦â€¦â€¦. + n.(n + 1) = [(n(n + 1)(n + 2))/3]

**Answer**

Let P(n): 1.2 + 2.3 + 3.4 + â€¦â€¦.. + n.(n + 1) = [(n(n+1)(n+2))/3]

For n = 1, LHS = 1.2= 2 and

RHS = [(1(n+1)(n + 2))/3] = (1.2.3)/3 = 1.2 = 2

We assume that P(n) is true for n = k

i.e., P(k): 1.2 + 2.3 + 3.4 + â€¦â€¦. +k(k + 1)

= [(k(k + 1)(k + 2))/3]Â Â Â â€¦â€¦.(i)

We have to prove that P(k +1) is true whenever P(k) is true.

âˆ´ P(k + 1) = [1.2 + 2.3 + 3.4 + â€¦â€¦. + k(k + 1)]Â + (k + 1)(k + 2)

= (k(k + 1)(k + 2))/3Â + (k + 1)(k + 2) [from (i)]

= (k(k + 1)(k + 2) + 3(k + 1)(k + 2))/3

= ((k + 1)(k + 2)(k + 3))/3

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by principle of mathematical induction P(n) is true for all values of n âˆˆ N.

7. 1.3 + 3.5 + 5.7 + â€¦â€¦â€¦. + (2n â€“ 1)(2n + 1) = (n(4n

^{2}Â + 6n â€“ 1))/3

**Answer**

Let P(n) be the given statement

i.e., P(n): 1.3 + 3.5 + 5.7+ â€¦.. + (2n â€“ 1)(2n + 1)

= (n(4n

^{2}Â + 6n â€“ 1))/3

Putting n = 1, LHS = 1. 3 = 3 and

RHS = (1(4.1

^{2}Â + 6.1 â€“ 1))/3 = (4 + 6 â€“ 1)/3 = 9/3 = 3

LHS = RHS

P(n) is true for n = 1

Assume that P(n) is true for n = k

i.e. P(k): 1.3 + 3.5 + 5.7 + â€¦ + (2k â€“ 1)(2k + 1)

= (k(4k

^{2}Â + 6k â€“ 1))/3 â€¦â€¦â€¦â€¦(i)

Now we have to prove that P(k + 1) is true whenever P(k) is true.

P(k + 1) = 1.3 + 3.5 + â€¦.. + (2k â€“ 1)(2k + 1) + {2(k + 1) â€“ 1}{2(k + 1) + 1}

= (k(4k

^{2}Â + 6k â€“ 1))/3 + (2k + 1)(2k + 3) [from (i)]

= ((4k

^{3}Â + 6k

^{2}Â â€“ k) + 3(2k + 1)(2k + 3))/3

= (4k

^{3}Â + 18k

^{2}Â + 23k + 9)/3

=((k + 1)(4k

^{2}Â + 14k + 9))/3

= ((k + 1)(4(k + 1)

^{2}Â + 6(k + 1) â€“ 1))/3

Thus, P(n) is true for n = k + 1

P(k + 1) is true whenever P(k) is true

Hence, by principle of mathematical induction

P(n) is true for all nÂ âˆˆ N

8. 1.2 + 2.2

^{2}Â + 3.2

^{2}Â + .. + n.2

^{n}Â = (n â€“ 1)2

^{n+1}Â + 2

**Answer**

Let P(n) be the given statement

i.e P(n): 1.2 + 2.2

^{2}Â + 3.2

^{2}+ .. + n. 2

^{n}

=(n â€“ 1)2

^{n+1}Â + 2

Putting n = 1, LHS = 1.2 =2; RHS = 0 + 2 = 2

P(n) is true for n = 1

Assume that P(n) is true for n = k i.e., P(k) is true

i.e., 1.2 + 2.2

^{2Â }+ 3.2

^{2}+â€¦. + k.2

^{k}Â = (k -1)2

^{k+1}Â + 2

We have to prove that P(k + 1) is true whenever P(k) is true.

âˆ´Â P(k + 1)

= 1.2 + 2.2

^{2}Â + 3.2

^{2}Â + â€¦ + k.2

^{k}Â + (k + 1)2

^{k+1}

= (k â€“ 1)2

^{k+1}Â + 2 + (k + 1)2

^{k+1}

= 2

^{k+1}Â [k -1 + k + 1] + 2 = 2k.2

^{k+1}Â + 2

=k 2

^{k+2}Â + 2

This proves P(n) true for n = k + 1

Thus P(k + 1) is true whenever P(k) is true.

Hence. P(n) is true for all nÂ âˆˆÂ N

9. 1/2 + 1/4 + 1/8 + â€¦â€¦â€¦ + 1/2

^{n}Â = 1 â€“ 1/2

^{n}

**Answer**

Let P(n): 1/2 + 1/4 + 1/8 + â€¦.. + 1/2

^{n}Â = 1 â€“ 1/2

^{n}

Pitting n = 1, LHS = 1/2, RHS = 1 â€“ 1/2 =1/2

i.e., LHS = RHS =1/2

P(n) is true for n = 1

Suppose P(n) is true for n = k

i.e., P(k): 1/2 + 1/4 + 1/8 + â€¦â€¦â€¦â€¦ + 1/2

^{K}Â = 1 â€“ 1/2

^{K}â€¦â€¦ (i)

we have to prove that P(k + 1) is true whenever P(k +1) is true.

P(k): 1/2 + 1/4 + 1/8 + â€¦.. + 1/2

^{K}Â + 1/2

^{K+ 1}

= 1 â€“ 1/2

^{K}Â + 1/2

^{K+1}[from (i)]

= 1 â€“ 1/2

^{K}(1 â€“ 1/2) = 1 â€“ 1/2

^{K}. 1/2 = 1 â€“1/2

^{K + 1}

This shows P(n) is true for n = k + 1

Thus P(k +1 ) is true whenever P(k) is true.

Hence, P(n) is true for all nÂ âˆˆÂ N

10.Â 1/2.5 + 1/5.8 + 1/8.11 + â€¦ + 1/((3n â€“ 1)(3n + 2)) = n/(6n + 4)

**Answer**

Let the given statement be P(n) i.e.

P(n):

1/2.5 + 1/5.8 + 1/8.11 + â€¦ + 1/((3n â€“ 1)(3n + 2)) = n/(6n + 4)

Putting n = 1, LHS = 1/2.5 = 1/10 and

RHS = 1/(6 + 4) = 1/10

âˆ´Â P(n) is true for n = 1

Let P(n) is true for n = k

i.e P(k): 1/2.5 + 1/5.8 + 1/8.11 + â€¦â€¦.. + 1/((3k â€“ 1)(3k + 2)) = k/(6k + 4) â€¦â€¦(i)

We have to prove that P(k + 1) is true whenever P(k) is true.

âˆ´Â P(k + 1) = 1/2.5 + 1/5.8 + 1/8.11 + â€¦â€¦.. + 1/((3k â€“ 1)(3k + 2)) + 1/((3k + 2)(3k + 5))

= k/(6k + 4) + 1/((3k + 2)(3k + 5)) [using (i)]

= (k(3k + 5) + 2)/(2(3k + 2)(3k + 5)) = (3k

^{2}Â + 5k + 2)/(2(3k + 2)(3k + 5))

= ((3k + 2)(k + 1))/(2(3k + 2)(3k + 5)) = (k + 1)/(2(3k + 5))

= (k + 1)/(6k + 10) = (k + 1)/(6(k + 1) + 4)

This shows that P(n) is true for n = k + 1

âˆ´Â P(k + 1) is true whenever P(k) is true

i.e., By principle of mathematical induction

P(n) is true for all nÂ âˆˆ N

11.Â 1/(1.2.3) + 1/(2.3.4) + 1/(3.4.5) + â€¦â€¦â€¦â€¦.+ 1/(n(n + 1)(n + 2)) = (n(n + 3))/(4(n + 1)(n + 2))

**Answer**

Let P(n): 1/(1.2.3) + 1/(2.3.4) + 1/(3.4.5) + â€¦â€¦â€¦â€¦.+ 1/(n(n + 1)(n + 2)) = (n(n + 3))/(4(n + 1)(n + 2))

Putting n = 1, LHS = 1/(1.2.3) = 1/6;

RHS = (1(1+3))/(4(1+1)(1+2)) = 4/(4.2.3) = 1/6

âˆ´Â LHS = RHS âˆ´ P(n) is true for n = 1

Assuming P(n) is true for n = k

i.e., P(k); 1/(1.2.3) + 1/(2.3.4) + 1/(3.4.5) + â€¦â€¦â€¦â€¦.. + 1/(k(k + 1)(k + 2)) = (k(k + 3))/(4(k +1)(k + 2)) â€¦(i)

We have to prove that P(k + 1) is true.

Whenever P(k) is true.

âˆ´Â P(k + 1): 1/(1.2.3) + 1/(2.3.4) + 1/(3.4.5) + â€¦â€¦.. + 1/(k(k + 1)(k + 2)) + 1/((k + 1)(k + 2)(k + 3))

= (k(k+3)/(4(k + 1)( k + 2)) + 1/((k + 1)(k + 2)( k + 3)) [from (i)]

= 1/((k + 1)(k + 2)) [(k(k + 3))/ 4 + 1/(k + 3)]

= 1/((k + 1)(k + 2))Â Ã—Â (k(k + 3)

^{2}Â + 4)/(4(k + 3))

= 1/(4(k + 1)(k + 2)(k + 3)) [k(k

^{2}Â + 6k + 9) + 4]

= (k

^{3}Â + 6k

^{2}Â + 9k + 4)/(4(k + 1)(k + 2)(k + 3))

=(k + 1)(k

^{2}Â + 5k + 4))/(4(k + 1)(k + 2)(k + 3))

= ((k + 1)(k + 4))/(4(k + 2)(k + 3)) =

This shows P(n) is true for n = k + 1 i.e.,

P(k + 1) is true whenever P(k) is true

Hence P(n) is true for all nÂ âˆˆÂ N

12. a + ar + ar

^{2}+ â€¦..+ar

^{n-1}Â = (a(rx

^{n}Â â€“ 1)/(r â€“ 1)

**Answer**

Let P(n): a + ar + ar

^{2}Â + â€¦.. + ar

^{n-1}

= (a(1 â€“ r

^{n}))/(1 â€“ r) , r 1 â€¦..(i)

For n = 1, LHS = a

And RHS = (a(1 â€“ r)/(1 â€“ r) = a

LHS = RHS i.e., P(i) is true.

Let us suppose that P(k) is true.

Putting n = k in (i), we have,

Â a + ar + ar

^{2}Â + â€¦..ar

^{k-1}= (a(1 â€“ r

^{k})/(1 â€“ r) + ar

^{k}â€¦..(ii)

Changing k to k + 1 in the last term arx

^{k-1}Â or LHS of (ii), it becomes ar

^{k+1-1}Â i.e., ar

^{k}.

Â Adding ar

^{k}Â to both sides of (ii), we have

a + ar + ar

^{2}Â + â€¦.+ ar

^{k-1}Â + ar

^{k}

= (a(1 â€“ r

^{k}))/(1 â€“ r) + ar

^{k}

= (a(1 â€“ r

^{k}) + ar

^{k}( 1 â€“ r))/(1 â€“ r)

= (a â€“ ar

^{k}Â + ar

^{k}Â â€“ ar

^{k+1})/(1 â€“ r)

= (a â€“ ar

^{k+1})/(1 â€“ r) = (a(1 â€“ r

^{k-1}))/(1 â€“ r) â€¦..(iii)

âˆ´Â P(n) is true for n = k + 1 i.e., P(k + 1) is true.

âˆ´Â By principle of mathematical induction, P(n) is true for natural numbers n.

13.Â (1 + 3/1)(1 + 5/4)(1 + 7/9) â€¦â€¦.(1 + (2n + 1)/n

^{2}) = (n + 1)

^{2}

**Answer**

Let the given statement be denoted by P(n) i.e. P(n):

Â (1 + 3/1)(1 + 5/4)(1 + 7/9) â€¦â€¦.(1 + (2n + 1)/n

^{2}) = (n + 1)

^{2}

Putting n = 1, LHS = 1 + (2Â Ã—Â 1 +1)/1

^{2}Â = 1 + 3 = 4

RHS = (1 + 1)

^{2}Â = 4

LHS = RHS

P(n) is true for n = 1

Suppose P(k) is true.

(1 + 3/1)(1 + 5/4)(1 + 7/9) â€¦.(1 + (2k + 1)/k

^{2})

= (k + 1)

^{2}

Kth factor = 1 + (2k + 1)/k

^{2}

(K + 1)th factor = 1 + (2(k + 1)+1)/(k +1)

^{2}Â = 1 + (2k + 3)/(k + 1)

^{2}

Multiplying both sides by 1 + (2k + 3)/(k + 1)

^{2}

LHS = (1 + 3/1)(1 + 5/4)(1 + 7/9)â€¦(1 + (2k+1)/k

^{2})(1 + (2k + 3)/(k + 1)

^{2})

RHS = (k + 1)

^{2}(1 + (2k + 3)/(k + 1)

^{2})

= (k + 1)

^{2}[((k + 1)

^{2}Â + 2k + 3)/(k + 1)

^{2}] = k

^{2}Â + 4k + 4

= (k + 2)

^{2}Â =

P(n) is true for n = k + 1 i.e., P(k + 1) is true

Whenever P(k) is true.

Hence, by principle of mathematical induction,

P(n) is true for all nÂ âˆˆÂ N

14.Â (1 + 1/1)(1 + 1/2)(1 + 1/3)â€¦â€¦(1 + 1/n) = (n + 1)

**Answer**

Let the given statements be P(n) i.e.,

P(n):

(1 + 1/1)(1 + 1/2)(1 + 1/3) â€¦â€¦.(1 + 1/n) = (n + 1)

For n = 1, LHS = ( 1 + 1/1) = 2 and RHS = 1+ 1 = 2

âˆ´ P(n) is true for n = 1

Let P(k) is true

i.e. P(k):

(1 + 1/1)(1 + 1/2)â€¦â€¦.. (1 + 1/k) = (k + 1)Â Â â€¦(i)

We have to prove that P(k + 1) is true.

Whenever P(k) is true

âˆ´ P(k + 1)

= [(1 + 1/1)(1 + 1/2)(1 + 1/3) â€¦â€¦â€¦(1 + 1/k)](1 + 1/(k +1)

= (k + 1)(1 + 1/(k + 1) = (k + 1)((k + 1+ 1)/(k + 1)) = k + 2

=Â

âˆ´ P(k + 1) is also true whenever P(k) is true.

Hence, by principle of mathematical induction

P(n) is true for all n âˆˆ N

15. 1

^{2}Â + 3

^{2}Â + 5

^{2}Â + â€¦â€¦.. + (2n â€“ 1)

^{2}Â = (n(2n -1)(2n +1))/3

**Answer**

Let the given statement be P(n) i.e.,

P(n): 1

^{2}Â + 3

^{2}Â + 5

^{2}Â + â€¦â€¦+(2n â€“ 1)

^{2}

=(n(2n -1)(2n + 1))/3

For n = 1, LHS = 1

^{2}Â = 1

RHS = (1.(2 â€“ 1)(2 + 1))/3 = (1.1.3)/3 = 1

P(n) is true for n = 1

Suppose P(k) is true for n = k i.e.,

1

^{2}Â + 3

^{2}Â + 5

^{2Â }+â€¦â€¦+(2k-1)

^{2}Â = (k(2k â€“ 1)(2k + 1))/3

Kth term = (2k -1)

^{2}

(k + 1)th term = (2(k +1) -1)

^{2}Â = (2k + 1)

^{2}

Adding (2k + 1)

^{2}Â to both sides.

LHS = 1

^{2}Â + 3

^{2}Â + 5

^{2}Â +â€¦.+(2k -1)

^{2}Â +(2k +1)

^{2}

RHS = (k(2k -1)(2k +1))/3 +(2k +1)

^{2}

= (2k +1)[(k(2k -1))/3 + (2k + 1)

= (2k +1)[(k(2k -1) + 3(2k +1))/3]

=(2k + 1)((2k

^{2}Â + 5k + 3)/3)

=((2k +1)(k + 1)(2k + 3))/3

=

Thus P(k + 1) is true for n = k i.e.,

P(k + 1) is true whenever P(k) is true

By principle of mathematical induction P(n) is true for all values of n âˆˆ N.

16. 1

^{2}Â + 3

^{2}Â + 5

^{2}Â + â€¦â€¦.. + (2n â€“ 1)

^{2}Â = (n(2n -1)(2n +1))/3

**Answer**

Let the given statement be P(n) i.e.,

P(n):

1/1.4 + 1/4.7 + 1/7.10 + â€¦â€¦+ 1/((3n -2)(3n +1)) = n/(3n +1)

For n = 1, LHS = 1/1.4 = 1/4 and RHS

= 1/(3.1 + 1) = 1/4

âˆ´Â P(n) is true for n = 1

Assume P(k) is true for n = k

âˆ´Â 1/1.4 + 1/4.7 + 1/7.10 + â€¦.+ 1/((3k -2)(3k + 1))

= k/(3k + 1) â€¦â€¦(i)

Now we have to prove that P(k + 1) is true whenever P(k) true.

P(k + 1): 1/1.4 + 1/4.7 + 1/7.10 + â€¦â€¦â€¦.. + 1/((3k -2)(3k + 1)) + 1/((3k +1)(3k + 4))

= k/(3k + 1) + 1/((3k + 1)(3k + 4))

= 1/(3k + 1) [k + 1/(3k + 4)] [from (i)]

= 1/(3k + 1) [(k(3k + 4)+1)/(3k + 4)] = (3k

^{2}Â + 4k +1)/((3k + 1)(3k + 4))

= ((k + 1)(3k + 1))/((3k + 1)(3k + 4)) = (k + 1)/(3k + 4) =Â

âˆ´Â P(k + 1) is true for n = (k + 1) i.e., P(k + 1) is true whenever P(k) is true.

Hence, by principle of mathematical induction

P(n) is true for all n âˆˆ N.

17.Â 1/3.5 + 1/5.7 + 1/7.9 + â€¦â€¦â€¦..+ 1/((2n + 1)(2n + 3)) = n/(3(2n + 3))

**Answer**

Let P(n) be the given statement i.e.

P(n): 1/3.5 + 1/5.7 + 1/7.9 + â€¦..+ 1/((2n + 1)(2n + 3)) = n/(3(2n + 3))

For n = 1, LHS = 1/3.5 = 1/15

RHS = 1/(3.(2 + 3)) = 1/3.5 = 1/15

Suppose P(k) be true for n = k i.e.

1/3.5 + 1/5.7 + 1/7.9+ â€¦.+ 1/((2k + 1)(2k + 3)) = k/(3(2k + 3))

Kth term = 1/((2k + 1)(2k + 3))

(k + 1)th term = 1/([2(k + 1)+1][2(k + 1) + 3]) = 1/((2k + 3)(2k + 5))

Adding 1/((2k + 3)(2k + 5)) to both sides,

LHS = 1/3.5 + 1/5.7 + 1/7.9 + â€¦â€¦â€¦+ 1/((2k + 1)(2k + 3)) + 1/((2k + 3)(2k + 5))

RHS = k/(3(2k + 3) + 1/((2k + 3)(2k + 5))

= 1/(2k + 3) [k/3 + 1/(2k + 5)]

= 1/(2k + 3) [(k(2k + 5) + 3)/(3(2k + 5))]

= (2k

^{2}Â + 5k + 3)/(3(2k + 3)(2k + 5)) = ((k + 1)(2k + 3))/(3(2k + 3)(2k + 5))

=Â

Hence P(k + 1) is true for n = k + 1, i.e., P(k + 1) is true whenever P(k) is true

Hence, by principle of mathematical induction

P(n) is true for all n âˆˆ N.

18. 1 + 2 + 3 + â€¦â€¦â€¦.+ n < 1/8 (2n + 1)

^{2}

**Answer**

Let P(n): 1 + 2 + 3 + â€¦â€¦ + n < 1/8 (2n + 1)

^{2}â€¦â€¦(i)

For n = 1, (i) becomes

1 < 1/8 (2 + 1)

^{2}Â => 1 < 9/8 => 1 < 1.1/8 which is true.

i.e., P(1) is tru.

Let us suppose that P(k) is true

i.e., 1 + 2+ 3 + â€¦.. + k < 1/8 (2k + 1)

^{2}â€¦..(ii)

We have to prove that P(k +1) is true.

Whenever P(k) is true.

Consider equation (ii)

Add k + 1 on both the sides,

1 + 2 + 3 + â€¦â€¦â€¦+ k + (k + 1) < 1/8 (2k + 1)

^{2}Â + k + 1

=> [1 + 2 + 3 â€¦â€¦â€¦ + k] + (k + 1) < (4k

^{2}Â + 4k + 1 + 8k + 8)/8

=> [1 + 2+ 3 + â€¦â€¦..+ k] + (k + 1) < 1/8 (4k

^{2}Â + 12k + 9)

=> 1 + 2 + 3 + â€¦â€¦+ k + (k + 1) < 1/8 (2k + 3)

^{2}

=> 1 + 2+ 3 + â€¦â€¦â€¦+ k +(k + 1)< 1/8[2(k + 1)+1]

^{2}

âˆ´Â P(k + 1) is true. Hence P(n) is true for all n (natural numbers).

19.Â n(n + 1)(n + 5) is a multiple of 3.

**Answer**

Let P(n): n(n + 1) (n + 5) is a multiple of 3

For n = 1, n(n + 1)(n + 5) = 1. 2. 6 = 12 = 3.4

Hence, P(n) is true for n = 1

Suppose P(k) is true for n = k

i.e. k(k + 1)(k + 5) = 3m

or k

^{3}Â + 6k

^{2}Â + 5k = 3m â€¦..(i)

Prove that P(k + 1) is true whenever P(k) is true.

P(k + 1): (k + 1)(k + 2)(k + 6)

= k(k

^{2}Â + 8k + 12) + (k

^{2}Â + 8k + 12)

= k

^{3}Â + 9k

^{2}Â + 20k + 12

= (k

^{3}Â + 6k

^{2}Â + 5k) + (3k

^{2}Â + 15k + 12)

= 3m + 3k

^{2}Â + 15k + 12 [ from (i)]

= 3(m + k

^{2}Â + 5k + 4) is multiple of 3

i.e., (k + 1)(k + 2)(k + 6) is a multiple of 3

i.e., P(k + 1) is multiple of 3, if P(k) is a multiple of 3

i.e., P(k + 1) is true whenever P(k) is true.

Hence P(n) is true for all nÂ âˆˆ N.

20. 10

^{2n-1}Â + 1 is divisible by 11.

**Answer**

Let P(n): 10

^{2n-1}Â + 1 is divisible by 11 for every natural number n.

For n = 1, P(1) = 10

^{2n-1}Â + 1 = (10 + 1) = 11 which is divisible by 11

âˆ´Â P(1) is true.

Let P(k) is true for n = k

i.e., P(k): 10

^{2k - 1}Â + 1 = 11m â€¦..(i)

we have to prove that P(k + 1) is true.

10

^{2(k+1)-1}Â + 1

= 10

^{2k - 1}Â . 10

^{2}Â + 1

= 100. 10

^{2k-1}Â + 1 = 100(11m â€“ 1) + 1 [using (i)]

= 100Â Ã—Â 11m â€“ 100 + 1 = 100Â Ã— 11m â€“ 99

= 11 (100m â€“ 9), which is divisible by 11

[ 11 is a factor of RHS]

âˆ´Â 10

^{2(k + 1)-1}Â + 1 is divisible by 11

âˆ´Â P(n) is true for n = k + 1 i.e., P(k + 1) is true.

âˆ´Â By principle of mathematical induction,

P(n) is true for all natural numbers n.

21. x

^{2n}- y

^{2n}Â is divisible by x + y

**Answer**

Let the statement be P(n) i.e. P(n): x

^{2n}Â â€“ y

^{2n}Â is divisible by x + y â€¦â€¦â€¦â€¦.(i)

Putting n = 1, x

^{2n}Â â€“ y

^{2n}Â = x

^{2}Â â€“ y

^{2}Â = (x + y)(x â€“ y)

Which is divisible by x + y

=> P(n) is true for n = 1

Let P(k) be true i.e. x

^{2k}Â â€“ y

^{2k}Â is divisible by x + y

Or x

^{2k}Â â€“ y

^{2k}Â = m(x + y)

Or x

^{2k}Â = m(x + y) + y

^{2k}â€¦â€¦â€¦.(ii)

We have to prove that P(k + 1) is true whenever P(k) is true.

x

^{2(k + 1)}Â - yx

^{(k + 1)}Â = x

^{2k + 2}Â â€“ y

^{2k + 2}

= x

^{2}Â . x

^{2k}Â â€“ y

^{2k + 2}

Putting the value of x

^{2k}Â from (ii)

= x

^{2}[m(x + y) + y

^{2k}] â€“ y

^{2k + 2}

= m(x + y) x

^{2}Â + x

^{2}y

^{2k}Â â€“ y

^{2k + 2}

=m(x + y)x

^{2}Â + y

^{2k}(x

^{2}Â â€“ y

^{2})

= m(x + y)x

^{2}Â + (x + y)(x â€“ y)y

^{2k}

= (x + y)[mx

^{2}Â + (x- y)y

^{2k}]

âˆ´Â x

^{2(k + 1)}Â â€“ y

^{2(k + 1)}Â is divisible by (x + y)

i.e., P(k + 1) is true whenever P(k) is true

Hence P(n) is true for all n âˆˆ N.

22. 3

^{2n+2}Â â€“ 8n â€“ 9 is divisible by 8.

**Answer**

Let P(n): 3

^{2n+2}Â â€“ 8n â€“ 9 is divisible by 8

For n = 1, 3

^{2n+2}Â â€“ 8n â€“ 9 = 3

^{2+2}Â â€“ 8.1 â€“ 9

=3

^{4}Â â€“ 8 â€“ 9 = 81 = 17

= 64 which is divisible by 8

Let P(k) be true.

i.e., P(k): 3

^{2k+2}Â â€“ 8k â€“ 9 = 8m, mÂ âˆˆÂ N

or 3

^{2k+2}Â = 8m + 8k + 9 â€¦..(i)

we have to prove that P(k + 1) is true whenever P(k) is true.

i.e., P(k + 1): 3

^{2(k + 1)+2}Â â€“ 8(k + 1) â€“ 9

= 3

^{2k+2+2}Â â€“ 8k â€“ 8 â€“ 9

= 3

^{2}.3

^{2k+2}Â â€“ 8k â€“ 17

= 9.3

^{2k+2}Â â€“ 8k â€“ 17

Putting the values of 3

^{2k+2}Â from (i)

= 9(8m + 8k + 9) â€“ 8k â€“ 17

= 72m + 72k + 81 â€“ 8k â€“ 17

= 72m + 64k + 64 = 8(9m + 8k + 8)

Hence, 3

^{2k+4}Â â€“ 8(k + 1) â€“ 9 is divisible by 8

i.e. P(k + 1) is true whenever P(k) is true.

Hence, P(n) is true for nÂ âˆˆ N.

23. 41

^{n}Â â€“ 14

^{n}Â is a multiple of 27.

**Answer**

Let P(n): 41

^{n}Â â€“ 14

^{n}Â is a multiple of 27

For n = 1, 41

^{n}Â â€“ 14

^{n}Â = 41 â€“ 14 = 27

âˆ´Â P(n) is true for n = 1

Let P(k) be true

i.e., P(k): 41

^{k}Â â€“ 14

^{k}Â is a multiple of 27.

i.e., P(k) = 41

^{k}Â â€“ 14

^{k}Â = 27m, mÂ âˆˆ N

or 41

^{k}Â = 27m + 14

^{k}â€¦â€¦(i)

We have to prove that P(k + 1) is true whenever P(k) is true

âˆ´Â P(k + 1): 41

^{k+1}Â â€“ 14

^{k+1}Â = 41.41

^{k}Â â€“ 14

^{k+1}

= 41(27m + 14

^{k}) â€“ 14

^{k+1}[from(i)]

= 41.27m + 41.14

^{k}Â â€“ 14

^{k+1}

= 27.41 m + 14

^{k}(41 â€“ 14)

= 27.41m + 14

^{k}. 27 = 27[41m + 14

^{k}]

This shows 41

^{k+1}Â â€“ 14

^{k+1}Â is a multiple of 27 or

P(k + 1) is true whenever P(k) is true

Hence, P(n) is true for all nÂ âˆˆÂ N.

24. (2n + 7)<(n + 3)

^{2}

**Answer**

Let P(n): (2n + 7)<(n + 3)

^{2}â€¦â€¦â€¦(i)

For n = 1, (i) becomes 2Â Ã—1 + 7 < ( 1+ 3)

^{2}

=> 9 < 16, which is true

P(1) is true

Let us suppose that P(k) is true

i.e., P(k): (2k + 7)< (k + 3)

^{2}â€¦â€¦..(ii)

now, we will prove that P(k + 1) is true

whenever P(k)is true i.e.,

P(k + 1): 2(k + 1) + 7 < (k + 4)

^{2}

We know that 2k + 7 < (k + 3)

^{2}[ from (ii)]

2k + 7 + 2 < (k + 3)

^{2}Â +2

[ adding 2 on both sides]

=> 2(k + 1) + 7 < (k + 3)

^{2}Â + 2 â€¦..(iii)

=> 2(k + 1)+ 7< k

^{2}Â + 6k + 11

Adding 2k + 5 to RHS

=> 2(k + 1) + 7 < k

^{2}Â + 7k + 16

Or 2(k + 1) + 7 < (k + 4)

^{2}

Thus P(k + 1) is true. Hence P(n) is true for n = k + 1

Hence, P(n) is true for all natural numbers n.