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.

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

1. Prove the following by using the principle of mathematical induction for all n∈ N
1 + 3 + 32 + …………. + 3n-1 = (3n – 1)/2

Answer

Let P(n) be the given statement.
i.e., P(n) : 1 + 3 + 32 + …………….. + 3n-1 = (3n – 1)/2
Putting n = 1, P(1) = ( 3 – 1)/2 = 1
P(n) is true for n = 1
Assume that P(k) is true.
i.e., P(k): 1 + 3 + 32 + ……….. + 3k -1
= (3 – 1)/2 ….(i)
We shall prove that P(k + 1) is true whenever P(k) is true.
P(k + 1) : 1 + 3 + 32 + ………… + 3(k + 1) - 1
= (3k + 1 – 1)/2
[1 + 3 + 32 + …….. + 3k-1] + 3k = (3k – 1)/2 + 3k
[using (i)]
(3k – 1 + 2.3k)/2 = ((1 + 2)3k – 1)/2 = (3.3k -1)/2
= (3k+1 – 1)/2
P(k + 1) is also true whenever P(k) is true.
Hence, P(n) is true for all n ∈ N

2. 13 + 23 + 33 + ………. + n3 = ((n(n + 1))/2)2

Answer

Let P(n) : 13 + 23 + 33 + ………. + n3
= (n2(n + 1)2)/4 ……..(i)
For n = 1, L.H.S = 13 = 1
and RHS = (12 (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) : 13 + 23 + 33 + ………… + k3 = (k2(k + 1)2)/4 …..(ii)
We have to prove that P(k + 1) is true whenever
P(k) is true.
P(k + 1) : 13 + 23+ 33 + ………. + k3 + (k + 1)3
= (k2 (k + 1)2)/4 + (k + 1)3 [using (i)]
= (k + 1)2[(k2 + 4(k + 1))/4]
= ((k + 1)2(k2 + 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) + ……+ 1/( 1+2+3+.....+n) = 2n/n +1

Answer

Let P(n): 1 + 1/(1 + 2) + 1/(1 + 2+ 3) + …….. + 1/(1 + 2+ 3+ …… + n) = 2n/n + 1 …. (i)
Putting n = 1,
LHS = 1 and RHS = (2.1)/(1+1) = 2/2 = 1
LHS = RHS ∴ P(1) is true
Let P(k) be true.
Putting n = k then
P(k): 1 + 1/(1 + 2) + …………. + 1/(1 + 2+ 3 + ……. + k)
=2k/k + 1 ……..(ii)
Now we shall prove that P(k + 1) is true whenever P(k) is true.
P(k + 1):
1 + 1/(1+2) + ………. + 1/(1 + 2 + 3 + …..+ k) + 1/(1 + 2 + 3 +……..(k + 1))
= 2k/(k + ) + 1/(1 + 2+ 3 + …(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 n ∈ N

4. 1.2.3 + 2.3.4 + ……….. + n(n + 1)(n + 2) =(n(n + 1)(n + 2)(n + 3))/4

Answer

Let P(n) : 1.2.3 + 2.3.4 + ………… + n(n+ 1)(n + 2) = (n(n +1)(n + 2)(n + 3))/4          ……….(i)
For n = 1, LHS = 1.2.3 = 6 and
RHS = (1(1 +1)(1 + 2)(1 + 3))/4 = (1 × 2 × 3 × 4)/4 = 6
LHS = RHS i.e., P(1) is true.
Let P(k) is true.
P(k): 1.2.3 + 2.3.4 + ……..+ k(k + 1)(k + 2) = (k(k + 1)(k + 2)(k + 3)/4   ………..(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 + …………. + k(k + 1)(k + 2)]+(k + 1)(k + 2)(k + 3)
= (k(k + 1)(k + 2)(k + 3)/4  + (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))/4          ……(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 + 2.32 + 3.33 + ………+ n.3n = ((2n -1)3n+1 + 3))/4

Answer

Let P(n): 1.3 + 2.32 + 3.33 + …….+ n.3n = ((2n – 1)3n+1 + 3)/4
Putting n = 1, P(1): LHS = 1.3 = 3
RHS = ((2 -1)32 + 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 + 2.32 + 3.33 + …… + k.3k = ((2k -1)3k+1 + 3)/4 ……..(i)
we have to prove that P(k + 1) is true whenever P(k) is true.
P(k + 1): 1.3 + 2.32 + ….. +(k + 1)3k+1 = [1.3 + 2.32 + 3.33 + ……. + k.3k] +(k + 1)3k + 1
= ((2k – 1).3k+1 + 3)/4 + (k + 1).3k+1 [using (i)]
=((2k – 1)3k+1 + 3 + 4(k + 1)3k+1)/4
= (3k+1 [2k – 1 + 4(k + 1)] + 3)/4
= (3k+1(6k + 3) + 3)/4 = ((2k + 1)3k+2 + 3)/4
=([2(k + 1)- 1]3(k+1)+1 + 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 n ∈ N

6. 1.2 + 2.3 + 3.4+ ………. + n.(n + 1) = [(n(n + 1)(n + 2))/3]

Answer

Let P(n): 1.2 + 2.3 + 3.4 + …….. + 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 + ……. +k(k + 1)
= [(k(k + 1)(k + 2))/3]      …….(i)
We have to prove that P(k +1) is true whenever P(k) is true.
∴ P(k + 1) = [1.2 + 2.3 + 3.4 + ……. + k(k + 1)]  + (k + 1)(k + 2)
= (k(k + 1)(k + 2))/3  + (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 ∈ N.

7. 1.3 + 3.5 + 5.7 + ………. + (2n – 1)(2n + 1) = (n(4n2 + 6n – 1))/3

Answer

Let P(n) be the given statement
i.e., P(n): 1.3 + 3.5 + 5.7+ ….. + (2n – 1)(2n + 1)
= (n(4n2 + 6n – 1))/3
Putting n = 1, LHS = 1. 3 = 3 and
RHS = (1(4.12 + 6.1 – 1))/3 = (4 + 6 – 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 + … + (2k – 1)(2k + 1)
= (k(4k2 + 6k – 1))/3 …………(i)
Now we have to prove that P(k + 1) is true whenever P(k) is true.
P(k + 1) = 1.3 + 3.5 + ….. + (2k – 1)(2k + 1) + {2(k + 1) – 1}{2(k + 1) + 1}
= (k(4k2 + 6k – 1))/3 + (2k + 1)(2k + 3) [from (i)]
= ((4k3 + 6k2 – k) + 3(2k + 1)(2k + 3))/3
= (4k3 + 18k2 + 23k + 9)/3
=((k + 1)(4k2 + 14k + 9))/3
= ((k + 1)(4(k + 1)2 + 6(k + 1) – 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 n ∈ N

8. 1.2 + 2.22 + 3.22 + .. + n.2n = (n – 1)2n+1 + 2

Answer

Let P(n) be the given statement
i.e P(n): 1.2 + 2.22 + 3.22+ .. + n. 2n
=(n – 1)2n+1 + 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+ 3.22+…. + k.2k = (k -1)2k+1 + 2
We have to prove that P(k + 1) is true whenever P(k) is true.
∴ P(k + 1)
= 1.2 + 2.22 + 3.22 + … + k.2k + (k + 1)2k+1
= (k – 1)2k+1 + 2 + (k + 1)2k+1
= 2k+1 [k -1 + k + 1] + 2 = 2k.2k+1 + 2
=k 2k+2 + 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 n ∈ N

9. 1/2 + 1/4 + 1/8 + ……… + 1/2n = 1 – 1/2n

Answer

Let P(n): 1/2 + 1/4 + 1/8 + ….. + 1/2n = 1 – 1/2n
Pitting n = 1, LHS = 1/2, RHS = 1 – 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 + ………… + 1/2K = 1 – 1/2K …… (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 + ….. + 1/2K + 1/2K+ 1
= 1 – 1/2K + 1/2K+1 [from (i)]
= 1 – 1/2K (1 – 1/2) = 1 – 1/2K . 1/2 = 1 –1/2K + 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 n ∈ N

10. 1/2.5 + 1/5.8 + 1/8.11 + … + 1/((3n – 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 + … + 1/((3n – 1)(3n + 2)) = n/(6n + 4)
Putting n = 1, LHS = 1/2.5 = 1/10 and
RHS = 1/(6 + 4) = 1/10
∴ 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 + …….. + 1/((3k – 1)(3k + 2)) = k/(6k + 4) ……(i)
We have to prove that P(k + 1) is true whenever P(k) is true.
∴ P(k + 1) = 1/2.5 + 1/5.8 + 1/8.11 + …….. + 1/((3k – 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)) = (3k2 + 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
∴ P(k + 1) is true whenever P(k) is true
i.e., By principle of mathematical induction
P(n) is true for all n ∈ N

11. 1/(1.2.3) + 1/(2.3.4) + 1/(3.4.5) + ………….+ 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) + ………….+ 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
∴ LHS = RHS ∴ 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) + ………….. + 1/(k(k + 1)(k + 2)) = (k(k + 3))/(4(k +1)(k + 2)) …(i)
We have to prove that P(k + 1) is true.
Whenever P(k) is true.
∴ P(k + 1): 1/(1.2.3) + 1/(2.3.4) + 1/(3.4.5) + …….. + 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)) × (k(k + 3)2 + 4)/(4(k + 3))
= 1/(4(k + 1)(k + 2)(k + 3)) [k(k2 + 6k + 9) + 4]
= (k3 + 6k2 + 9k + 4)/(4(k + 1)(k + 2)(k + 3))
=(k + 1)(k2 + 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 n ∈ N

12. a + ar + ar2+ …..+arn-1 = (a(rxn – 1)/(r – 1)

Answer

Let P(n): a + ar + ar2 + ….. + arn-1
= (a(1 – rn))/(1 – r) , r 1 …..(i)
For n = 1, LHS = a
And RHS = (a(1 – r)/(1 – 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,
 a + ar + ar2 + …..ark-1= (a(1 – rk)/(1 – r) + ark …..(ii)
Changing k to k + 1 in the last term arxk-1 or LHS of (ii), it becomes ark+1-1 i.e., ark.
 Adding ark to both sides of (ii), we have
a + ar + ar2 + ….+ ark-1 + ark
= (a(1 – rk))/(1 – r) + ark
= (a(1 – rk) + ark( 1 – r))/(1 – r)
= (a – ark + ark – ark+1)/(1 – r)
= (a – ark+1)/(1 – r) = (a(1 – rk-1))/(1 – r) …..(iii)
∴ 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 natural numbers n.

13. (1 + 3/1)(1 + 5/4)(1 + 7/9) …….(1 + (2n + 1)/n2) = (n + 1)2

Answer

Let the given statement be denoted by P(n) i.e. P(n):
 (1 + 3/1)(1 + 5/4)(1 + 7/9) …….(1 + (2n + 1)/n2) = (n + 1)2
Putting n = 1, LHS = 1 + (2 × 1 +1)/12 = 1 + 3 = 4
RHS = (1 + 1)2 = 4
LHS = RHS
P(n) is true for n = 1
Suppose P(k) is true.
(1 + 3/1)(1 + 5/4)(1 + 7/9) ….(1 + (2k + 1)/k2)
= (k + 1)2
Kth factor = 1 + (2k + 1)/k2
(K + 1)th factor = 1 + (2(k + 1)+1)/(k +1)2 = 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)…(1 + (2k+1)/k2)(1 + (2k + 3)/(k + 1)2)
RHS = (k + 1)2(1 + (2k + 3)/(k + 1)2)
= (k + 1)2[((k + 1)2 + 2k + 3)/(k + 1)2] = k2 + 4k + 4
= (k + 2)2 =
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 n ∈ N

14. (1 + 1/1)(1 + 1/2)(1 + 1/3)……(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) …….(1 + 1/n) = (n + 1)
For n = 1, LHS = ( 1 + 1/1) = 2 and RHS = 1+ 1 = 2
∴ P(n) is true for n = 1
Let P(k) is true
i.e. P(k):
(1 + 1/1)(1 + 1/2)…….. (1 + 1/k) = (k + 1)   …(i)
We have to prove that P(k + 1) is true.
Whenever P(k) is true
∴ P(k + 1)
= [(1 + 1/1)(1 + 1/2)(1 + 1/3) ………(1 + 1/k)](1 + 1/(k +1)
= (k + 1)(1 + 1/(k + 1) = (k + 1)((k + 1+ 1)/(k + 1)) = k + 2

∴ P(k + 1) is also true whenever P(k) is true.
Hence, by principle of mathematical induction
P(n) is true for all n ∈ N

15. 12 + 32 + 52 + …….. + (2n – 1)2 = (n(2n -1)(2n +1))/3

Answer

Let the given statement be P(n) i.e.,
P(n): 12 + 32 + 52 + ……+(2n – 1)2
=(n(2n -1)(2n + 1))/3
For n = 1, LHS = 12 = 1
RHS = (1.(2 – 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.,
12 + 32 + 5+……+(2k-1)2 = (k(2k – 1)(2k + 1))/3
Kth term = (2k -1)2
(k + 1)th term = (2(k +1) -1)2 = (2k + 1)2
Adding (2k + 1)2 to both sides.
LHS = 12 + 32 + 52 +….+(2k -1)2 +(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)((2k2 + 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 ∈ N.

16. 12 + 32 + 52 + …….. + (2n – 1)2 = (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 + ……+ 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
∴ P(n) is true for n = 1
Assume P(k) is true for n = k
∴ 1/1.4 + 1/4.7 + 1/7.10 + ….+ 1/((3k -2)(3k + 1))
= k/(3k + 1) ……(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 + ……….. + 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)] = (3k2 + 4k +1)/((3k + 1)(3k + 4))
= ((k + 1)(3k + 1))/((3k + 1)(3k + 4)) = (k + 1)/(3k + 4) = 
∴ 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 ∈ N.

17. 1/3.5 + 1/5.7 + 1/7.9 + ………..+ 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 + …..+ 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+ ….+ 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 + ………+ 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))]
= (2k2 + 5k + 3)/(3(2k + 3)(2k + 5)) = ((k + 1)(2k + 3))/(3(2k + 3)(2k + 5))

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 ∈ N.

18. 1 + 2 + 3 + ……….+ n < 1/8 (2n + 1)2

Answer

Let P(n): 1 + 2 + 3 + …… + n < 1/8 (2n + 1)2 ……(i)
For n = 1, (i) becomes
1 < 1/8 (2 + 1)2 => 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 + ….. + k < 1/8 (2k + 1)2 …..(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 + ………+ k + (k + 1) < 1/8 (2k + 1)2 + k + 1
=> [1 + 2 + 3 ……… + k] + (k + 1) < (4k2 + 4k + 1 + 8k + 8)/8
=> [1 + 2+ 3 + ……..+ k] + (k + 1) < 1/8 (4k2 + 12k + 9)
=> 1 + 2 + 3 + ……+ k + (k + 1) < 1/8 (2k + 3)2
=> 1 + 2+ 3 + ………+ k +(k + 1)< 1/8[2(k + 1)+1]2
∴ P(k + 1) is true. Hence P(n) is true for all n (natural numbers).

19. 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 k3 + 6k2 + 5k = 3m …..(i)
Prove that P(k + 1) is true whenever P(k) is true.
P(k + 1): (k + 1)(k + 2)(k + 6)
= k(k2 + 8k + 12) + (k2 + 8k + 12)
= k3 + 9k2 + 20k + 12
= (k3 + 6k2 + 5k) + (3k2 + 15k + 12)
= 3m + 3k2 + 15k + 12 [ from (i)]
= 3(m + k2 + 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 n ∈ N.

20. 102n-1 + 1 is divisible by 11.

Answer

Let P(n): 102n-1 + 1 is divisible by 11 for every natural number n.
For n = 1, P(1) = 102n-1 + 1 = (10 + 1) = 11 which is divisible by 11
∴ P(1) is true.
Let P(k) is true for n = k
i.e., P(k): 102k - 1 + 1 = 11m …..(i)
we have to prove that P(k + 1) is true.
102(k+1)-1 + 1
= 102k - 1 . 102 + 1
= 100. 102k-1 + 1 = 100(11m – 1) + 1 [using (i)]
= 100 × 11m – 100 + 1 = 100 × 11m – 99
= 11 (100m – 9), which is divisible by 11
[ 11 is a factor of RHS]
∴ 102(k + 1)-1 + 1 is divisible by 11
∴ 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.

21. x2n - y2n is divisible by x + y

Answer

Let the statement be P(n) i.e. P(n): x2n – y2n is divisible by x + y ………….(i)
Putting n = 1, x2n – y2n = x2 – y2 = (x + y)(x – y)
Which is divisible by x + y
=> P(n) is true for n = 1
Let P(k) be true i.e. x2k – y2k is divisible by x + y
Or x2k – y2k = m(x + y)
Or x2k = m(x + y) + y2k ……….(ii)
We have to prove that P(k + 1) is true whenever P(k) is true.
x2(k + 1) - yx(k + 1) = x2k + 2 – y2k + 2
= x2 . x2k – y2k + 2
Putting the value of x2k from (ii)
= x2[m(x + y) + y2k] – y2k + 2
= m(x + y) x2 + x2y2k – y2k + 2
=m(x + y)x2 + y2k(x2 – y2)
= m(x + y)x2 + (x + y)(x – y)y2k
= (x + y)[mx2 + (x- y)y2k]
∴ x2(k + 1) – y2(k + 1) 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 ∈ N.

22. 32n+2 – 8n – 9 is divisible by 8.

Answer

Let P(n): 32n+2 – 8n – 9 is divisible by 8
For n = 1, 32n+2 – 8n – 9 = 32+2 – 8.1 – 9
=34 – 8 – 9 = 81 = 17
= 64 which is divisible by 8
Let P(k) be true.
i.e., P(k): 32k+2 – 8k – 9 = 8m, m ∈ N
or 32k+2 = 8m + 8k + 9 …..(i)
we have to prove that P(k + 1) is true whenever P(k) is true.
i.e., P(k + 1): 32(k + 1)+2 – 8(k + 1) – 9
= 32k+2+2 – 8k – 8 – 9
= 32.32k+2 – 8k – 17
= 9.32k+2 – 8k – 17
Putting the values of 32k+2 from (i)
= 9(8m + 8k + 9) – 8k – 17
= 72m + 72k + 81 – 8k – 17
= 72m + 64k + 64 = 8(9m + 8k + 8)
Hence, 32k+4 – 8(k + 1) – 9 is divisible by 8
i.e. P(k + 1) is true whenever P(k) is true.
Hence, P(n) is true for n ∈ N.

23. 41n – 14n is a multiple of 27.

Answer

Let P(n): 41n – 14n is a multiple of 27
For n = 1, 41n – 14n = 41 – 14 = 27
∴ P(n) is true for n = 1
Let P(k) be true
i.e., P(k): 41k – 14k is a multiple of 27.
i.e., P(k) = 41k – 14k = 27m, m ∈ N
or 41k = 27m + 14k ……(i)
We have to prove that P(k + 1) is true whenever P(k) is true
∴ P(k + 1): 41k+1 – 14k+1 = 41.41k – 14k+1
= 41(27m + 14k) – 14k+1 [from(i)]
= 41.27m + 41.14k – 14k+1
= 27.41 m + 14k(41 – 14)
= 27.41m + 14k. 27 = 27[41m + 14k]
This shows 41k+1 – 14k+1 is a multiple of 27 or
P(k + 1) is true whenever P(k) is true
Hence, P(n) is true for all n ∈ N.

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

Answer

Let P(n): (2n + 7)<(n + 3)2 ………(i)
For n = 1, (i) becomes 2 ×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 ……..(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 +2
[ adding 2 on both sides]
=> 2(k + 1) + 7 < (k + 3)2 + 2 …..(iii)
=> 2(k + 1)+ 7< k2 + 6k + 11
Adding 2k + 5 to RHS
=> 2(k + 1) + 7 < k2 + 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.
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