If P_{m} stands for ^{m}P_{m' }then the value of 1 + P_{1} + 2P_{2} + 3P_{3} + ..... + nP_{n} is
n!
n^{2}
( n + 1 )!
( n - 1 ) !
A student was asked to prove a statement P (n) by method of induction. He proved that P (3 ) is true such that P (n) = P (n + 1 ) for all
n ∈ N
n ≥ 3
n ∈ I
n < 3
The remainder, when number 5^{99} is divided by 13, is
2
8
12
32
P (n) = P (n + 1 ) for all natural numbers n, then P (n) is true ?
For all n
For all n > 1
For all n > m
Nothing can be said
The solution of the inequality is.
( ^{2}/_{3}, 8 )
( -2, ^{8}/_{3} )
If n > 1 and x ≠ 0. then expression ( 1 + x)^{n} - nx -1 is divisible by
x^{2}
x^{3}
x^{5}
x^{7}
The number 101 x 102 x 103 x 104 x ..... x 107 is divisible by .
4000
4050
5040
5050
By the mathematical induction, the expression 11^{n+2} + 12^{2n + 1} is divisible by
133
124
114
113
The unit digit in the number 7^{126} is
1
3
9
5
If x > -1, then the statement ( 1 + x )^{ n} > 1 + nx is true for
All n < 1
All n > 1
All n ∈ N
All n > 1 provided x ≠ 0