maihuna wrote:Find the sum of :

1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + ...+ 12x13x14

7250
8190
8930
9250
9660

Each term in the sum is a product of three consecutive integers, so each term is a multiple of 3! = 6. Since we're adding multiples of 6, the sum is a multiple of 6 (and particularly of 3). Adding the digits in each answer, only 8190 and 9660 could be correct.

Now, 8190 is not divisible by 4, while 9660 is, so if we can decide whether the sum is a multiple of 4, we'll know which answer choice is the right one. In our sum, which I'll call S, most of the terms are multiples of 4, and will add to some multiple of 4 -- call it 4k. Now,

S = 4k + 1*2*3 + 5*6*7 + 9*10*11
S = 2(2k + 1*3 + 5*3*7 + 9*5*11)
S = 2(even + odd + odd + odd)
S = 2(some odd number)

So S is a multiple of 2, but not a multiple of 4, and 9660 cannot be correct, and the answer must be 8190.

sureshbala wrote: The method suggested by Ian is absolutely fine, but if there is one more option such that it leaves a remainder 2 when divided by 4 and 0 when divided by 3 we need to calculate the sum which can be done as follows...

Suresh has a good general solution above, but I wanted to suggest an approach that did not use formulas (sum of the first n cubes, for example) which are never required for the GMAT. The question in the original post is not a realistic GMAT question, since it would be too time consuming for most. There are other ways to solve it besides the two approaches above; by grouping the terms and factoring appropriately, you can get down to small numbers fairly easily without using any special formulas, but those solutions aren't especially interesting, and do take a bit of time to complete.