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Principle of Mathematical Induction Class 11 MCQs Questions with Answers

Question 1.
The sum of the series 1³ + 2³ + 3³ + ………..n³ is
(a) {(n + 1)/2}²
(b) {n/2}²
(c) n(n + 1)/2
(d) {n(n + 1)/2}²

Answer

Answer: (d) {n(n + 1)/2}²
Hint:
Given, series is 1³ + 2³ + 3³ + ……….. n³
Sum = {n(n + 1)/2}²


Question 2.
If n is an odd positive integer, then an + bn is divisible by :
(a) a² + b²
(b) a + b
(c) a – b
(d) none of these

Answer

Answer: (b) a + b
Hint:
Given number = an + bn
Let n = 1, 3, 5, ……..
an + bn = a + b
an + bn = a³ + b³ = (a + b) × (a² + b² + ab) and so on.
Since, all these numbers are divisible by (a + b) for n = 1, 3, 5,…..
So, the given number is divisible by (a + b)


Question 3.
1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ….. + 1/{n(n + 1)}
(a) n(n + 1)
(b) n/(n + 1)
(c) 2n/(n + 1)
(d) 3n/(n + 1)

Answer

Answer: (b) n/(n + 1)
Hint:
Let the given statement be P(n). Then,
P(n): 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ….. + 1/{n(n + 1)} = n/(n + 1).
Putting n = 1 in the given statement, we get
LHS = 1/(1 ∙ 2) = and RHS = 1/(1 + 1) = 1/2.
LHS = RHS.
Thus, P(1) is true.
Let P(k) be true. Then,
P(k): 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ….. + 1/{k(k + 1)} = k/(k + 1) ..…(i)
Now 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ….. + 1/{k(k + 1)} + 1/{(k + 1)(k + 2)}
[1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ….. + 1/{k(k + 1)}] + 1/{(k + 1)(k + 2)}
= k/(k + 1)+1/{ (k + 1)(k + 2)}.
{k(k + 2) + 1}/{(k + 1)²/[(k + 1)k + 2)] using …(ii)
= {k(k + 2) + 1}/{(k + 1)(k + 2}
= {(k + 1)² }/{(k + 1)(k + 2)}
= (k + 1)/(k + 2) = (k + 1)/(k + 1 + 1)
⇒ P(k + 1): 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ……… + 1/{ k(k + 1)} + 1/{(k + 1)(k + 2)}
= (k + 1)/(k + 1 + 1)
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1)is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.


Question 4.
The sum of the series 1² + 2² + 3² + ………..n² is
(a) n(n + 1)(2n + 1)
(b) n(n + 1)(2n + 1)/2
(c) n(n + 1)(2n + 1)/3
(d) n(n + 1)(2n + 1)/6

Answer

Answer: (d) n(n + 1)(2n + 1)/6
Hint:
Given, series is 1² + 2² + 3² + ………..n²
Sum = n(n + 1)(2n + 1)/6


Question 5.
{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. {1 – 1/(n + 1)} =
(a) 1/(n + 1) for all n ∈ N.
(b) 1/(n + 1) for all n ∈ R
(c) n/(n + 1) for all n ∈ N.
(d) n/(n + 1) for all n ∈ R

Answer

Answer: (a) 1/(n + 1) for all n ∈ N.
Hint:
Let the given statement be P(n). Then,
P(n): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. {1 – 1/(n + 1)} = 1/(n + 1).
When n = 1, LHS = {1 – (1/2)} = ½ and RHS = 1/(1 + 1) = ½.
Therefore LHS = RHS.
Thus, P(1) is true.
Let P(k) be true. Then,
P(k): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 1)
Now, [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] ∙ [1 – {1/(k + 2)}]
= [1/(k + 1)] ∙ [{(k + 2 ) – 1}/(k + 2)}]
= [1/(k + 1)] ∙ [(k + 1)/(k + 2)]
= 1/(k + 2)
Therefore p(k + 1): [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 2)
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.


Question 6.
For any natural number n, 7n – 2n is divisible by
(a) 3
(b) 4
(c) 5
(d) 7

Answer

Answer: (c) 5
Hint:
Given, 7n – 2n
Let n = 1
7n – 2n = 71 – 21 = 7 – 2 = 5
which is divisible by 5
Let n = 2
7n – 2n = 72 – 22 = 49 – 4 = 45
which is divisible by 5
Let n = 3
7n – 2n = 73 – 23 = 343 – 8 = 335
which is divisible by 5
Hence, for any natural number n, 7n – 2n is divisible by 5


Question 7.
1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + …….. + 1/{n(n + 1)(n + 2)} =
(a) {n(n + 3)}/{4(n + 1)(n + 2)}
(b) (n + 3)/{4(n + 1)(n + 2)}
(c) n/{4(n + 1)(n + 2)}
(d) None of these

Answer

Answer: (a) {n(n + 3)}/{4(n + 1)(n + 2)}
Hint:
Let P (n): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……. + 1/{n(n + 1)(n + 2)} = {n(n + 3)}/{4(n + 1)(n + 2)} .
Putting n = 1 in the given statement, we get
LHS = 1/(1 ∙ 2 ∙ 3) = 1/6 and RHS = {1 × (1 + 3)}/[4 × (1 + 1)(1 + 2)] = ( 1 × 4)/(4 × 2 × 3) = 1/6.
Therefore LHS = RHS.
Thus, the given statement is true for n = 1, i.e., P(1) is true.
Let P(k) be true. Then,
P(k): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……… + 1/{k(k + 1)(k + 2)} = {k(k + 3)}/{4(k + 1)(k + 2)}. ……. (i)
Now, 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………….. + 1/{k(k + 1)(k + 2)} + 1/{(k + 1)(k + 2)(k + 3)}
= [1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………..…. + 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)}] [using(i)]
= {k(k + 3)² + 4}/{4(k + 1)(k + 2)(k + 3)}
= (k³ + 6k² + 9k + 4)/{4(k + 1)(k + 2)(k + 3)}
= {(k + 1)(k + 1)(k + 4)}/{4 (k + 1)(k + 2)(k + 3)}
= {(k + 1)(k + 4)}/{4(k + 2)(k + 3)
⇒ P(k + 1): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……….….. + 1/{(k + 1)(k + 2)(k + 3)}
= {(k + 1)(k + 2)}/{4(k + 2)(k + 3)}
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.


Question 8.
The nth terms of the series 3 + 7 + 13 + 21 +………. is
(a) 4n – 1
(b) n² + n + 1
(c) none of these
(d) n + 2

Answer

Answer: (b) n² + n + 1
Hint:
Let S = 3 + 7 + 13 + 21 +……….an-1 + an …………1
and S = 3 + 7 + 13 + 21 +……….an-1 + an …………2
Subtract equation 1 and 2, we get
S – S = 3 + (7 + 13 + 21 +……….an-1 + an) – (3 + 7 + 13 + 21 +……….an-1 + an)
⇒ 0 = 3 + (7 – 3) + (13 – 7) + (21 – 13) + ……….+ (an – an-1) – an
⇒ 0 = 3 + {4 + 6 + 8 + ……(n-1)terms} – an
⇒ an = 3 + {4 + 6 + 8 + ……(n-1)terms}
⇒ an = 3 + (n – 1)/2 × {2 ×4 + (n – 1 – 1)2}
⇒ an = 3 + (n – 1)/2 × {8 + (n – 2)2}
⇒ an = 3 + (n – 1) × {4 + n – 2}
⇒ an = 3 + (n – 1) × (n + 2)
⇒ an = 3 + n² + n – 2
⇒ an = n² + n + 1
So, the nth term is n² + n + 1


Question 9.
n(n + 1)(n + 5) is a multiple of ____ for all n ∈ N
(a) 2
(b) 3
(c) 5
(d) 7

Answer

Answer: (b) 3
Hint:
Let P(n) : n(n + 1)(n + 5) is a multiple of 3.
For n = 1, the given expression becomes (1 × 2 × 6) = 12, which is a multiple of 3.
So, the given statement is true for n = 1, i.e. P(1) is true.
Let P(k) be true. Then,
P(k) : k(k + 1)(k + 5) is a multiple of 3
⇒ K(k + 1)(k + 5) = 3m for some natural number m, … (i)
Now, (k + 1)(k + 2)(k + 6) = (k + 1)(k + 2)k + 6(k + 1)(k + 2)
= k(k + 1)(k + 2) + 6(k + 1)(k + 2)
= k(k + 1)(k + 5 – 3) + 6(k + 1)(k + 2)
= k(k + 1)(k + 5) – 3k(k + 1) + 6(k + 1)(k + 2)
= k(k + 1)(k + 5) + 3(k + 1)(k +4) [on simplification]
= 3m + 3(k + 1 )(k + 4) [using (i)]
= 3[m + (k + 1)(k + 4)], which is a multiple of 3
⇒ P(k + 1) : (k + 1 )(k + 2)(k + 6) is a multiple of 3
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.


Question 10.
Find the number of shots arranged in a complete pyramid the base of which is an equilateral triangle, each side containing n shots.
(a) n(n+1)(n+2)/3
(b) n(n+1)(n+2)/6
(c) n(n+2)/6
(d) (n+1)(n+2)/6

Answer

Answer: (b) n(n+1)(n+2)/6
Hint:
Let each side of the base contains n shots,
then the number of shots in the lowest layer = n + (n – 1) + (n – 2) + ………..+ 1
= n(n + 1)/2
= (n² + n)/2
Now, write (n – 1), (n – 2), ….. for n, then we obtain the number of shots in 2nd, 3rd…layers
So, Total shots = ∑(n² + n)/2
= (1/2)×{∑n² + ∑n}
= (1/2)×{n(n+1)(2n+1)/6 + n(n+1)/2}
= n(n+1)(n+2)/6


Question 11.
For any natural number n, 7n – 2n is divisible by
(a) 3
(b) 4
(c) 5
(d) 7

Answer

Answer: (c) 5
Hint:
Given, 7n – 2n
Let n = 1
7n – 2n = 71 – 21 = 7 – 2 = 5
which is divisible by 5
Let n = 2
7n – 2n = 72 – 22 = 49 – 4 = 45
which is divisible by 5
Let n = 3
7n – 2n = 7³ – 2³ = 343 – 8 = 335
which is divisible by 5
Hence, for any natural number n, 7n – 2n is divisible by 5


Question 12.
(n² + n) is ____ for all n ∈ N.
(a) Even
(b) odd
(c) Either even or odd
(d) None of these

Answer

Answer: (a) Even
Hint:
Let P(n): (n² + n) is even.
For n = 1, the given expression becomes (1² + 1) = 2, which is even.
So, the given statement is true for n = 1, i.e., P(1)is true.
Let P(k) be true. Then,
P(k): (k² + k) is even
⇒ (k² + k) = 2m for some natural number m. ….. (i)
Now, (k + 1)² + (k + 1) = k² + 3k + 2
= (k² + k) + 2(k + 1)
= 2m + 2(k + 1) [using (i)]
= 2[m + (k + 1)], which is clearly even.
Therefore, P(k + 1): (k + 1)² + (k + 1) is even
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n)is true for all n ∈ N.


Question 13.
For all n ∈ N, 3×52n+1 + 23n+1 is divisible by
(a) 19
(b) 17
(c) 23
(d) 25

Answer

Answer: (b) 17
Hint:
Given, 3 × 52n+1 + 23n+1
Let n = 1,
3 × 52×1+1 + 23×1+1 = 3 × 52+1 + 23+1 = 3 × 5³ + 24 = 3 × 125 + 16 = 375 + 16 = 391
Which is divisible by 17
Let n = 2,
3 × 52×2+1 + 23×2+1 = 3 × 54+1 + 26+1 = 3 × 55 + 27 = 3 × 3125 + 128 = 9375 + 128
= 9503
Which is divisible by 17
Hence, For all n ∈ N, 3 × 52n+1 + 23n+1 is divisible by 17


Question 14.
Find the number of shots arranged in a complete pyramid the base of which is an equilateral triangle, each side containing n shots.
(a) n(n+1)(n+2)/3
(b) n(n+1)(n+2)/6
(c) n(n+2)/6
(d) (n+1)(n+2)/6

Answer

Answer: (b) n(n+1)(n+2)/6
Hint:
Let each side of the base contains n shots,
then the number of shots in the lowest layer = n + (n – 1) + (n – 2) + ………..+ 1
= n(n + 1)/2
= (n² + n)/2
Now, write (n – 1), (n – 2), ….. for n, then we obtain the number of shots in 2nd, 3rd…layers
So, Total shots = ∑(n² + n)/2
= (1/2) × {∑n² + ∑n}
= (1/2) × {n(n+1)(2n+1)/6 + n(n+1)/2}
= n(n+1)(n+2)/6


Question 15.
{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. {1 – 1/(n + 1)} =
(a) 1/(n + 1) for all n ∈ N.
(b) 1/(n + 1) for all n ∈ R
(c) n/(n + 1) for all n ∈ N.
(d) n/(n + 1) for all n ∈ R

Answer

Answer: (a) 1/(n + 1) for all n ∈ N.
Hint:
Let the given statement be P(n). Then,
P(n): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. {1 – 1/(n + 1)} = 1/(n + 1).
When n = 1, LHS = {1 – (1/2)} = ½ and RHS = 1/(1 + 1) = ½.
Therefore LHS = RHS.
Thus, P(1) is true.
Let P(k) be true. Then,
P(k): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 1)
Now, [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] ∙ [1 – {1/(k + 2)}]
= [1/(k + 1)] ∙ [{(k + 2 ) – 1}/(k + 2)}]
= [1/(k + 1)] ∙ [(k + 1)/(k + 2)]
= 1/(k + 2)
Therefore p(k + 1): [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 2)
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.


Question 16.
(1 + x)n ≥ ____ for all n ∈ N,where x > -1
(a) 1 + nx
(b) 1 – nx
(c) 1 + nx/2
(d) 1 – nx/2

Answer

Answer: (a) 1 + nx
Hint:
Let P(n): (1 + x) )n ≥ (1 + nx).
For n = 1, we have LHS = (1 + x))1 = (1 + x), and
RHS = (1 + 1 ∙ x) = (1 + x).
Therefore LHS ≥ RHS is true.
Thus, P(1) is true.
Let P(k) is true. Then,
P(k): (1 + x)1 ≥ (1 + kx). …….. (i)
Now,(1 + x)k+1 = (1 + x)k (1 + x)
≥ (1 + kx)(1 + x) [using (i)]
=1 + (k + 1)x + kx²
≥ 1 + (k + 1)x + x [Since kx² ≥ 0]
Therefore P(k + 1) : (1 + x)k + 1 ≥ 1 + (k + 1)x
⇒ P(k +1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.


Question 17.
102n-1 + 1 is divisible by ____ for all N ∈ N
(a) 9
(b) 10
(c) 11
(d) 13

Answer

Answer: (c) 11
Hint:
Let P (n): (102n-1 + 1) is divisible by 11.
For n=1, the given expression becomes {10(2×1-1) + 1} = 11, which is divisible by 11.
So, the given statement is true for n = 1, i.e., P (1) is true.
Let P(k) be true. Then,
P(k): (102k-1 + 1) is divisible by 11
⇒ (102k-1 + 1) = 11 m for some natural number m.
Now, {102(k-1)-1 – 1 + 1} = (102k+1 + 1) = {10² ∙ 10(2k+1)+ 1}
= 100 × {102k-1 + 1 } – 99
= (100 × 11 m) – 99
= 11 × (100 m – 9), which is divisible by 11
⇒ P (k + 1) : {102(k-1) – 1 + 1} is divisible by 11
⇒ P (k + 1) is true, whenever P(k) is true.
Thus, P (1) is true and P(k + 1) is true , whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.


Question 18.
For all n∈N, 72n − 48n−1 is divisible by :
(a) 25
(b) 2304
(c) 1234
(d) 26

Answer

Answer: (b) 2304
Hint:
Given number = 72n − 48n − 1
Let n = 1, 2 ,3, 4, ……..
72n − 48n − 1 = 7² − 48 − 1 = 49 – 48 – 1 = 49 – 49 = 0
72n − 48n − 1 = 74 − 48 × 2 − 1 = 2401 – 96 – 1 = 2401 – 97 = 2304
72n − 48n − 1 = 76 − 48 × 3 − 1 = 117649 – 144 – 1 = 117649 – 145 = 117504 = 2304 × 51
Since, all these numbers are divisible by 2304 for n = 1, 2, 3,…..
So, the given number is divisible by 2304


Question 19.
The sum of the series 1² + 2² + 3² + ………..n² is
(a) n(n + 1)(2n + 1)
(b) n(n + 1)(2n + 1)/2
(c) n(n + 1)(2n + 1)/3
(d) n(n + 1)(2n + 1)/6

Answer

Answer: (d) n(n + 1)(2n + 1)/6
Hint:
Given, series is 1² + 2² + 3² + ………..n²
Sum = n(n + 1)(2n + 1)/6


Question 20.
{1/(3 ∙ 5)} + {1/(5 ∙ 7)} + {1/(7 ∙ 9)} + ……. + 1/{(2n + 1)(2n + 3)} =
(a) n/(2n + 3)
(b) n/{2(2n + 3)}
(c) n/{3(2n + 3)}
(d) n/{4(2n + 3)}

Answer

Answer: (c) n/{3(2n + 3)}
Hint:
Let the given statement be P(n). Then,
P(n): {1/(3 ∙ 5) + 1/(5 ∙ 7) + 1/(7 ∙ 9) + ……. + 1/{(2n + 1)(2n + 3)} = n/{3(2n + 3).
Putting n = 1 in the given statement, we get
and LHS = 1/(3 ∙ 5) = 1/15 and RHS = 1/{3(2 × 1 + 3)} = 1/15.
LHS = RHS
Thus, P(1) is true.
Let P(k) be true. Then,
P(k): {1/(3 ∙ 5) + 1/(5 ∙ 7) + 1/(7 ∙ 9) + …….. + 1/{(2k + 1)(2k + 3)} = k/{3(2k + 3)} ….. (i)
Now, 1/(3 ∙ 5) + 1/(5 ∙ 7) + ..…… + 1/[(2k + 1)(2k + 3)] + 1/[{2(k + 1) + 1}2(k + 1) + 3
= {1/(3 ∙ 5) + 1/(5 ∙ 7) + ……. + [1/(2k + 1)(2k + 3)]} + 1/{(2k + 3)(2k + 5)}
= k/[3(2k + 3)] + 1/[2k + 3)(2k + 5)] [using (i)]
= {k(2k + 5) + 3}/{3(2k + 3)(2k + 5)}
= (2k² + 5k + 3)/[3(2k + 3)(2k + 5)]
= {(k + 1)(2k + 3)}/{3(2k + 3)(2k + 5)}
= (k + 1)/{3(2k + 5)}
= (k + 1)/[3{2(k + 1) + 3}]
= P(k + 1) : 1/(3 ∙ 5) + 1/(5 ∙ 7) + …….. + 1/[2k + 1)(2k + 3)] + 1/[{2(k + 1) + 1}{2(k + 1) + 3}]
= (k + 1)/{3{2(k + 1) + 3}]
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for n ∈ N.


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