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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, 7^{n} – 2^{n} is divisible by

(a) 3

(b) 4

(c) 5

(d) 7

## Answer

Answer: (c) 5

Hint:

Given, 7^{n} – 2^{n}

Let n = 1

7^{n} – 2^{n} = 7^{1} – 2^{1} = 7 – 2 = 5

which is divisible by 5

Let n = 2

7^{n} – 2^{n} = 7^{2} – 2^{2} = 49 – 4 = 45

which is divisible by 5

Let n = 3

7^{n} – 2^{n} = 7^{3} – 2^{3} = 343 – 8 = 335

which is divisible by 5

Hence, for any natural number n, 7^{n} – 2^{n} 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 +……….a_{n-1} + a_{n} …………1

and S = 3 + 7 + 13 + 21 +……….a_{n-1} + a_{n} …………2

Subtract equation 1 and 2, we get

S – S = 3 + (7 + 13 + 21 +……….a_{n-1} + a_{n}) – (3 + 7 + 13 + 21 +……….a_{n-1} + a_{n})

⇒ 0 = 3 + (7 – 3) + (13 – 7) + (21 – 13) + ……….+ (a_{n} – a_{n-1}) – a_{n}

⇒ 0 = 3 + {4 + 6 + 8 + ……(n-1)terms} – a_{n}

⇒ a_{n} = 3 + {4 + 6 + 8 + ……(n-1)terms}

⇒ a_{n} = 3 + (n – 1)/2 × {2 ×4 + (n – 1 – 1)2}

⇒ a_{n} = 3 + (n – 1)/2 × {8 + (n – 2)2}

⇒ a_{n} = 3 + (n – 1) × {4 + n – 2}

⇒ a_{n} = 3 + (n – 1) × (n + 2)

⇒ a_{n} = 3 + n² + n – 2

⇒ a_{n} = 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, 7^{n} – 2^{n} is divisible by

(a) 3

(b) 4

(c) 5

(d) 7

## Answer

Answer: (c) 5

Hint:

Given, 7^{n} – 2^{n}

Let n = 1

7^{n} – 2^{n} = 7^{1} – 2^{1} = 7 – 2 = 5

which is divisible by 5

Let n = 2

7^{n} – 2^{n} = 7^{2} – 2^{2} = 49 – 4 = 45

which is divisible by 5

Let n = 3

7^{n} – 2^{n} = 7³ – 2³ = 343 – 8 = 335

which is divisible by 5

Hence, for any natural number n, 7^{n} – 2^{n} 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, 3×5^{2n+1} + 2^{3n+1} is divisible by

(a) 19

(b) 17

(c) 23

(d) 25

## Answer

Answer: (b) 17

Hint:

Given, 3 × 5^{2n+1} + 2^{3n+1}

Let n = 1,

3 × 5^{2×1+1} + 2^{3×1+1} = 3 × 5^{2+1} + 2^{3+1} = 3 × 5³ + 24 = 3 × 125 + 16 = 375 + 16 = 391

Which is divisible by 17

Let n = 2,

3 × 5^{2×2+1} + 2^{3×2+1} = 3 × 5^{4+1} + 2^{6+1} = 3 × 5^{5} + 2^{7} = 3 × 3125 + 128 = 9375 + 128

= 9503

Which is divisible by 17

Hence, For all n ∈ N, 3 × 5^{2n+1} + 2^{3n+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.

10^{2n-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): (10^{2n-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): (10^{2k-1} + 1) is divisible by 11

⇒ (10^{2k-1} + 1) = 11 m for some natural number m.

Now, {10^{2(k-1)-1} – 1 + 1} = (10^{2k+1} + 1) = {10² ∙ 10^{(2k+1)}+ 1}

= 100 × {10^{2k-1} + 1 } – 99

= (100 × 11 m) – 99

= 11 × (100 m – 9), which is divisible by 11

⇒ P (k + 1) : {10^{2(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, 7^{2n} − 48n−1 is divisible by :

(a) 25

(b) 2304

(c) 1234

(d) 26

## Answer

Answer: (b) 2304

Hint:

Given number = 7^{2n} − 48n − 1

Let n = 1, 2 ,3, 4, ……..

7^{2n} − 48n − 1 = 7² − 48 − 1 = 49 – 48 – 1 = 49 – 49 = 0

7^{2n} − 48n − 1 = 7^{4} − 48 × 2 − 1 = 2401 – 96 – 1 = 2401 – 97 = 2304

7^{2n} − 48n − 1 = 7^{6} − 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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