The sum of n terms of the following series 13 + 33 + 53 + 73 + ... is
n2(2n2 - 1)
n3(n - 1)
n3 + 8n + 4
2n4 + 3n2
If w is an imginary cube root of unity, then the value of (2 - w)(2 - w2) + 2(3 - w)(3 - w2) + ... + (n - 1)(n - w)(n - w2) is
If the first and (2n - 1)th terms of an AP, GP and HP are equal and their nth terms are respectively a, b, c, then always
a = b = c
a + c = b
ac - b2 = 0
Let a, b, c and d be any four real numbers. Then, an +bn = cn + dn holds for any natural number n, if
a + b = c + d
a - b = c - d
a + b = c + d, a2 + b2 = c2 + d2
a - b = c - d, a2 - b2 = c2 - d2
Let x1, x2, ..., x15 be 15 distinct numbers chosen from 1, 2, 3, ..., 15. Then, the value of (x1 - 1)(x2 - 1)(x3 - 1)...(x15 - 1) is
always 0
0
always even
always odd
Let d(n) denotes the number of divisors of n including 1 and itself. Then, d (225), d (1125) and d(640) are
in AP
in HP
in GP
consecutive integers
Let S = , where N is the set of all natural numbers. Then, the number of elements in the set
6
7
13
14
If x and y are digits such that 17! = 3556xy428096000, then x + y equals
15
6
12
13
A.
15
Given, 17! = 3556xy428096000
Since. 17! is divisible by 9, so sum of the digits (48 + x + y) must be divisible by 9.
So, x + y can be 15 or 6
Also, 17! is divisible by 11, so [10 + x - y] must be multiple of 11 or 0. The only possibility is
Let f(x) = x + 1/2. Then, the number of real values of x for which the three unequal terms f(x), f(2x), f(4x) are in HP is
1
0
3
2