Mathematical Induction Iit Jee Questions Pdf

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

A.
676
B.
627
C.
547
D.
239

Solution:

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

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

A.
64
B.
36
C.
49
D.
none of these

Solution:

Which is divided by 49 by the case of above.

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

A.
64
B.
24
C.
17
D.
none of these

Solution:

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

A.
25
B.
26
C.
2309
D.
none of these

Solution:

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

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

A.
P (k) ⇒ P (k + 2)
B.
P (1) is correct
C.
Principal of mathematical induction can be used to prove the formula
D.
P (k) ⇒ P (k+1)

Solution:

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

A.
all n > 1
B.
all n
C.
all n > 3
D.
none of these

Solution:

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

A.
a + b
B.
(a+b)²
C.
a³+b³
D.
none of these.

Solution:

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

A.
2
B.
8
C.
6
D.
15

Solution:

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

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

A.
19
B.
21
C.
17
D.
23

Solution:

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

A.
16
B.
7
C.
8
D.
none of these.

Solution:

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

A.
32
B.
16
C.
19
D.
8

Solution:

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

A.
all n ≥ 3
B.
all n ≥ 2
C.
all n
D.
no n ∈ N

Solution:

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

A.
5
B.
6
C.
3
D.
none of these

Solution:

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

A.
3
B.
1
C.
5
D.
7

Solution:

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

A.
4
B.
3
C.
2
D.
none of thes

Solution:

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

A.
P (3) is not true
B.
P (5) is not true
C.
P (41) is not true
D.
P (1) is not true

Solution:

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

A.
P (n) is true for all n < m
B.
P (n) is true for all n ≥ m
C.
P (n) is true for all n
D.
none of these.

Solution:

Consider the statement P (n) : "n²≥ 100 ". Here P(n) ⇒ P(n+1) for all natural numbers n. Does it mean

A.
P (n) is true for all n ≥ 2
B.
P (n) is true for all n ≥ 3
C.
P (n) is true for all n.
D.
none of these

Solution:

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

A.
3
B.
2
C.
4
D.
1

Solution:

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 :

A.
2
B.
1
C.
3
D.
4

Solution:

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

A.
for all n > 5
B.
for all n < 5
C.
for all n ≥ 5
D.
none of these

Solution:

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

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 :

A.
a - b
B.
a+b
C.
a³+b³
D.
none of these

Solution:

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

A.
1/4 (n + 1) (n -1) (n + 2) (n + 3)
B.
1/4 (n + 1) (n + 2) (n + 3)
C.
1/4 n (n + 1) (n + 2) (n + 3)
D.
none of these

Solution:

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)

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

A.
1/2 (n) (n + 2)
B.
1/2 (n + 1) (n + 2)
C.
1/2 (n + 1) (n + 3)
D.
none of these

Solution:

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

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

A.
(n−1)³+n³
B.
(n+1)³+8n²
C.
D.
none of these

Solution: