rishianand7 wrote: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

Try to determine the last two digits of the sum.
Rephrase the values in terms of 40² and work from the INSIDE OUT:

40²
39² + 41² = (40-1)² + (40+1)² = (40² - 2*40*1 + 1) + (40² + 2*40*1 + 1) = 2*40² + 2.
38² + 42² = (40-2)² + (40+2)² = (40² - 2*40*2 + 4) + (40² + 2*40*2 + 4) = 2*40² + 8.
37² + 43² = (40-3)² + (40+3)² = (40² - 2*40*3 + 9) + (40² + 2*40*3 + 9) = 2*40² + 18.
36² + 44² = (40-4)² + (40+4)² = (40² - 2*40*4 + 16) + (40² + 2*40*4 + 16) = 2*40² + 32.

The last two digits of the sum will be equal to the sum of the values in red:
2+8+18+32 = 60.

The correct answer is C.

Another way is to use the formula for the sum of the first n perfect squares: 1/6 * n * (n+1) * (2n+1)

(Sum of the first 44 perfect squares) - (Sum of the first 35 perfect squares) = Sum we want

Simplify first:

(1/6 * 44 * 45 * 89) - (1/6 * 35 * 36 * 71)

Simplify further:

= 1/6 * (44*45*89 - 35*36*71)

= 1/6 * 180 * (11 * 89 - 7 * 71)

= 30 * (11 * 89 - 7 * 71)

= 30 * (979 - 497)

= 10 * 3 * (482)

= 10 * 1446

= 14460

is this a typical gmat question?

faraz_jeddah wrote:is this a typical gmat question?

The posted problem is beyond the scope of the GMAT.
But even problems beyond the scope of the GMAT can yield helpful take-aways.
One take-away in my solution above: to know which answer choice is correct, we need only determine the last two digits of the sum.
A useful strategy that can be applied to other problems.