laxieqv wrote:

I came across this one, a very very gripping maths question. You may put a finger on it

.

36^2 + 37^2 + 38^2 + 39^2 + 40^2 + 41^2 + 42^2 + 43^2 + 44^2 =

(A) 14400

(B) 14440

(C) 14460

(D) 14500

(E) 14520

See if anyone have other solutions. I will put my own if no one has the same way of solving.

Good one. This can be solved as follows..

36^2 + 37^2 + 38^2 + 39^2 + 40^2 + 41^2 + 42^2 + 43^2 + 44^2 =

(1^2 + 2^ 2 + .... + 44 ^2 ) - (1^2 + 2 ^2 + ... + 35^2)

The formula to compute sum of the squares of first N numbers is n(n+1)(2n+1)/6

Using this we get (1^2 + 2^ 2 + .... + 44 ^2 ) = (44*45*89)/6 = 29370

similarly (1^2 + 2 ^2 + ... + 35^2) = (35*36*71)/6 = 14910

Subtracting both we get 14460 - C

Thanks