Question

Class 11 Maths MCQ Questions of Principle of Mathematical Induction with Answers?

Answer 1

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Practice MCQ Questions for class 11 Maths Chapter-Wise

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

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

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)

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

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

6. For any natural number n, 7ⁿ – 2ⁿ is divisible by

(a) 3
(b) 4
(c) 5
(d) 7

7. 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

8. 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

9. For any natural number n, 7ⁿ – 2ⁿ is divisible by

(a) 3
(b) 4
(c) 5
(d) 7

10. For all n ∈ N, 3 × 5²ⁿ⁺¹ + 2³ⁿ⁺¹ is divisible by

(a) 19
(b) 17
(c) 23
(d) 25

11. For all n ∈ N, 7²ⁿ − 48n −1 is divisible by :

(a) 25
(b) 2304
(c) 1234
(d) 26

12. Let P(n) be the statement 2ⁿ <n! where n is a natural number, then P(n) is true for:

(a) all n
(b) all n>2
(c) all n>3
(d) all n<3

13. Let T(k) be the statement 1+3+5+....+(2k−1) = k²+10. Which of the following is correct?

(a) T(1) is true
(b) T(k) is true ⇒ T(k + 1) is true
(c) T(n) is true for all n ∈ N
(d) All above are correct

14. Let P(n)  :n²+n+1 is an even integer. If P(k) is assumed true ⇒ P(k+1) is true. Therefore, P(n) is

(a) true for n>1,n∈N
(b) true for all n∈N
(c) true for n>2,n∈N
(d) none of these

15. If x ⁿ⁻¹ is divisible by x−k, then the least positive integral value of k is

(a) 1
(b) 2
(c) 3
(d) 4

16. if n is a +ve integer, then 2.4²ⁿ⁺¹ + 3³ⁿ⁺¹is divisible by

(a) 2
(b) 7
(c) 11
(d) 27

17. A student was asked to prove a statement P(n) by induction. He proved that P(k + 1) is true whenever P(k) is true for all k > 5 ∈ N and also that P (5) is true. On the basis of this he could conclude that P(n) is true

(a) for all n ∈ N
(b) for all n > 5
(c) for all n ≥ 5
(d) for all n < 5

18.The greatest positive integer, which divides n(n +1)(n + 2)(n + 3) for all n ∈ N , is

(a) 2
(b) 6
(c) 24
(d) 120

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

(a) 3
(b) 8
(c) 5
(d) 7

20. For every positive integer n, 7n – 3n is divisible by

(a) 3
(b) 4
(c) 5
(d) 7

Answer 2

1. Answer: (d) {n(n + 1)/2}²

Explanation: Given, series is 1³ + 2³ + 3³ + ……….. n³

Sum = {n(n + 1)/2}²

2. Answer: (b) a + b

Explanation: Given number = an + bn

Let n = 1, 3, 5, ……..

a_{n +} b_n = a + b

a_n + b_n = 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)

3. Answer: (b) n/(n + 1)

Explanation: 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.

4. Answer: (d) n(n + 1)(2n + 1)/6

Explanation: Given, series is 1² + 2²+ 3² + ………..n²

Sum = n(n + 1)(2n + 1)/6

5. Answer: (a) 1/(n + 1) for all n ∈ N.

Explanation: 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)} = 1/2 and RHS = 1/(1 + 1) = 1/2.

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.

6. Answer: (c) 5

Explanation: 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.

7. Answer: (b) n² + n + 1

Explanation: Let S = 3 + 7 + 13 + 21 +……….a_n-1 + an …………1

and S = 3 + 7 + 13 + 21 +……….a_n-1 + an …………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} – an

⇒ 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

⇒ an = n² + n + 1

So, the nth term is n² + n + 1

8. Answer: (d) (n+1)(n+2)/6

Explanation: 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

= (n  + 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

Answer 3

9. Answer: (c) 5

Explanation: 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

10. Answer: (b) 17

Explanation: Given, 3 × 5²ⁿ⁺¹ + 2³ⁿ⁺¹

Let n = 1,

3 × 5^2×1+1 + 2^3×1+1 

= 3 × 5²⁺¹ + 2³⁺¹ 

= 3 × 5³ + 2⁴ 

= 3 × 125 + 16 

= 375 + 16 

= 391

Which is divisible by 17

Let n = 2,

3 × 5^2×2+1 + 2^3×2+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

11. Answer: (b) 2304

Explanation: 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

12. Answer: (c) all n>3

Explanation: We have,P(n) be the statement 2ⁿ <n!

Where n is a natural number

Then,

Put n=1,2,3....

So, P(1) be the statement of 2¹<1! = 2<1 (It is wrong)

P(2) be the statement of 2² < 2! = 4<2 (It is wrong)

P(3)  be the statement of  <3!=8<6(It is wrong)

But,

P(4)  be the statement of 2⁴ <4! =16<24(Itisright)

Similarly,

P(5), P(6)....... So, all number n is a natural number.

13. Answer: (b) T(k) is true ⇒ T(k + 1) is true

Explanation: When k = 1, LHS = 1 but RHS = 1 + 10 = 11

∴ T(1) is not true

Let T(k) is true.

i.e., 1+3+5+.....+(2k−1) = k²+10

Now, 1+ 3+ 5 + ..... + (2k -1) + (2k +1)

= k²+10+2k+1 = (k+1)²+10

∴ T(k +1) is true.

i.e., T(k) is true ⇒ T (k +1) is true.

But T(n) is not true for all n ∈ N , as T(1) is not true.

14. Answer: (d) none of these    

Explanation: Given, P(n)  :n²+n+1 is an even integer.

For n=1,P(1)=3, which is not an even integer.

∴P(1) is not true.

Also, n²+n+1 = n(n+1)+1 is always an odd integer. (principle of induction is not applicable).

15. Answer: (a) 1

Explanation: x⁻¹ is always a factor of xⁿ⁻¹.

⇒k =1.

16. Answer: (c) 11

Explanation: Let P(n) =2.4²ⁿ⁺¹ + 3³ⁿ⁺¹

Then P(1)=2.4³+3⁴ =209, which is divisible by 11 but not divisible by 2, 7 or 27.

Further, let P(k)=2.4^2k+1 + 3^3k+1 is divisible by 11, i.e., 2.4^2k+1+ 3^3k+1=11q for some integer q.

Now P(k+1)=2.4^2k+3 + 3^3k+4

=2.4^2k+1.4² + 3^3k+1.3³

=16.2.4^2k+1+27.3^3k+1

=16.2.2^2k+1+(16+11).3^3k+1

=16[2.4^2k+1+3^3k+1]+11.3^3k+1

=16.11q+11.3^3k+1

=11(16q+3^3k+1)

=11m

where m=16q+33k+1 is another integer.

∴p(k+1) is divisible by 11.

∴P(n)=2.4²ⁿ⁺¹+3³ⁿ⁺¹ is divisible by 11

for all n∈N.

17. Answer: (c) for all n ≥ 5

Explanation: Since P(5) is true and P(k + 1) is true, whenever P (k) is true.

18. Answer: (c) 24

Explanation: The product of r consecutive integers is divisible by r ! . Thus n (n+ 1 ) (n + 2) (n + 3) is divisible by 4 ! = 24.

19. Answer: (a) 3

Explanation: Let P(n) = n(n + 1)(n + 5)

Substituting n = 1, 2, 3,….

P(1) = 1(1 + 1)(1 + 5) = 2(6) = 12; multiple of 2, 3, 4, 6

P(2) = 2(2 + 1)(2 + 5) = 2(3)(7) = 42; multiple of 2, 3, 6, 7

P(3) = 3(3 + 1)(3 + 5) = 3(4)(8) = 96; multiple of 2, 3, 4, 6, 8, 12

20. Answer: (b) 4

Explanation: Let P(n) = 7ⁿ – 3ⁿ

Substituting n = 1, 2, 3,…

P(1) = 71 – 31 = 7 – 3 = 4

P(2) = 72 – 32 = 49 – 9 = 40

P(3) = 73 – 33 = 343 – 27 = 316

Thus, for every positive integer n, 7n – 3n is divisible by 4.

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