kirankp wrote:

The sum of the first n positive perfect squares, where n is a positive integer, is given by the formula \(\frac{n^3}{3} + c*n^2 + \frac{n}{6}\), where \(c\) is a constant. What is the sum of the first 15 positive perfect squares?

(A) 1,010

(B) 1,164

(C) 1,240

(D) 1,316

(E) 1,476

First of all there is a direct formula also provided above by bunuel i.e. [(n)(n+1)(2n+1)]/6

now if we do not know this and directly put 15 in place of N ..

It'll come >> 15[225/3+15c+1/6] = 15[(451+90c)/6] = 5[(451 + 90c)/2] ..

now (450 + 90c) should be an even integer so that it should get divisible by 2, that figured out c has to be in fraction and as (450 + 90c) is an even integer answer should have "0" in the last(because it'll be multiplied by "5" outside [ ] ) .. we can eliminate B,D,E ryt away ..

for choosing between A and C. I took 1/2 as my first no. and bingo I got the answer :D

Note that OA is C, not D. Check here:

Hope it helps.