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Test: Principle Of Mathematical Induction- 2 - JEE MCQ


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25 Questions MCQ Test Chapter-wise Tests for JEE Main & Advanced - Test: Principle Of Mathematical Induction- 2

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Test: Principle Of Mathematical Induction- 2 - Question 1

If n is a +ve integer, then 33n−26n−1 is divisible by

Detailed Solution for Test: Principle Of Mathematical Induction- 2 - Question 1

For n=1, 33n − 26n − 1
= 33 − 27 = 0
For n=2, 33n− 26n − 1
= 36 − 52 − 1
= 729 − 53
= 676

Test: Principle Of Mathematical Induction- 2 - Question 2

If n is a +ve integer, then 23n−7n−1 is divisible by

Detailed Solution for Test: Principle Of Mathematical Induction- 2 - Question 2


Which is divided by 49 by the case of above.

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Test: Principle Of Mathematical Induction- 2 - Question 3

If n is a +ve integer, then 3.52n+1+23n+1 is divisible by

Detailed Solution for Test: Principle Of Mathematical Induction- 2 - Question 3


Test: Principle Of Mathematical Induction- 2 - Question 4

If n is a +ve integer, then 72n−4 is divisible by

Detailed Solution for Test: Principle Of Mathematical Induction- 2 - Question 4

For n = 1, 72n – 4 = 72 – 4 = 49 – 4 = 45 
45 is not divisible by any of the options.

Test: Principle Of Mathematical Induction- 2 - Question 5

Let P (k) = 1 + 3 + 5 + …………….+ (2k – 1) = (3+k2).Then which of the following is true ?

Test: Principle Of Mathematical Induction- 2 - Question 6

The inequality 2n<n!,n∈N is true for :

Test: Principle Of Mathematical Induction- 2 - Question 7

For each n ∈ N,a2n−1+b2n−1 is divisible by :

Test: Principle Of Mathematical Induction- 2 - Question 8

For each n ∈ N, n (n + 1) (2n + 1) is divisible by :

Detailed Solution for Test: Principle Of Mathematical Induction- 2 - Question 8

n(n+1)(2n+1) is divisible by 6
Suppose P(n) : n(n+1)(2n+1) 
P(1) : 1(1+1)(2(1)+1)
= 2(3) = 6
The statement is true for P(1)
The statement is true for P(m)
P(m) : m(m+1)(2m+1) is divisible by 6
P(m+1) : [(m+1)(m+1+1)(2(m+1)+1]
= (m+1)(m+2)(2m+3)
= (m+1) m(2m+1+2) + 2(m+1)(2m+3)
= m(m+1)(2m+1) + 2m(m+1) + 2(m+1)(2m+3)
6k + 2(m+1)(m+2m+3)
6k + 2(m+1)(3m+3)
6k + 6(m+1)(m+1)
= 6[k + (m+1)2]
P(m+1) is true for m+1
It is also true for all n, 
n(n+1)(2n+1) is divisible by 6

Test: Principle Of Mathematical Induction- 2 - Question 9

For each n ∈ N, 3(52n+1)+23n+1 is divisible by :

Test: Principle Of Mathematical Induction- 2 - Question 10

For each n ∈ N, 23n−1 is divisible by :

Test: Principle Of Mathematical Induction- 2 - Question 11

For each n ∈ N, 32n−1 is divisible by :

Test: Principle Of Mathematical Induction- 2 - Question 12

The statement P (n) : “(n+3)> 2n+3 “ is true for :

Detailed Solution for Test: Principle Of Mathematical Induction- 2 - Question 12


Test: Principle Of Mathematical Induction- 2 - Question 13

The smallest positive integer ‘n‘ for which 2n(1×2×3×...............×n) < nn holds is :

Test: Principle Of Mathematical Induction- 2 - Question 14

If 10n+3×4n+1+k is divisible by 9 for all n ∈ N, then the least positive integral value of k is 

Test: Principle Of Mathematical Induction- 2 - Question 15

The smallest positive integer n, for which ( 1 × 2 × 3 ×……×n ) 

Test: Principle Of Mathematical Induction- 2 - Question 16

Let P(n) : n2−n+41 is a prime number . then :

Test: Principle Of Mathematical Induction- 2 - Question 17

Let that P(n) ⇒ P(n+1) for all natural numbers n. also, if P (m) is true, m ∈ N, then we conclude that

Test: Principle Of Mathematical Induction- 2 - Question 18

Consider the statement P (n) : “n≥ 100 “. Here P(n) ⇒ P(n+1) for all natural numbers n. Does it mean

Test: Principle Of Mathematical Induction- 2 - Question 19

The smallest positive integer ‘n‘ for which P (n) : 2< (1×2×3×............×n) holds is :

Test: Principle Of Mathematical Induction- 2 - Question 20

If xn−1 is divisible by x – k for all n belongs to natural numbers N, then the least positive integral value of k is :

Test: Principle Of Mathematical Induction- 2 - Question 21

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

Detailed Solution for Test: Principle Of Mathematical Induction- 2 - Question 21

If P(a) is true and truth of P(n) implies that P(n+1) is also true, then, we can conclude that P(n) is true for all n≥a where n∈N.
Here, P(5) is true.
Hence we can conclude that P(n) is true for all n≥5

Test: Principle Of Mathematical Induction- 2 - Question 22

If a,b,c ∈ N, an+bn is divisible by c, when n is odd but not when n is even, then the value of c is :

Test: Principle Of Mathematical Induction- 2 - Question 23

1.2.3 + 2.3.4 + 3.4.5 + ………..up to n terms is equal to :

Detailed Solution for Test: Principle Of Mathematical Induction- 2 - Question 23

nth term = n(n+1)(n+2); where n=1,2,3,…
= n(n2+3n+2) = n3 +3n2 +2n
Sum =n(n+1)/2 ; if nth term =n
= n(n+1)(2n+1)/6 ; if nth term =n2
= n2(n+1)2/4 ; if nth term =n3
Hence the required sum =
n2(n+1)2 /4 + 3 ×n(n+1)(2n+1)/6 +2 × n(n+1)/2
= n2 (n+1)2 /4 +n(n+1)(2n+1)/2 + n(n+1)
= n(n+1) {n(n+1)/4 + (2n+1)/2 +1}
= n(n+1) {(n2 + n + 4n + 2 + 4)/4}
= 1/4 n(n+1){ n2 + 5n + 6}
= 1/4 n(n+1)(n+2)(n+3)

Test: Principle Of Mathematical Induction- 2 - Question 24

The number of terms in the expansion of (x+y+z)n is

Detailed Solution for Test: Principle Of Mathematical Induction- 2 - Question 24

Number of terms is equal to  n+r-1Cn
​For (x+y+z)n
r = 3
Hence number of terms are n+3-1Cn
= n+2Cn
​= ((n+2)!)/2!n!
= [(n+1)(n+2)]/2

Test: Principle Of Mathematical Induction- 2 - Question 25

The sum of the terms in the nth bracket of the series 1 + (2+3+4) + (5+6+7+8+9) ….is

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