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.maihuna wrote:Find the sum of :
1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + ...+ 12x13x14
7250
8190
8930
9250
9660
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.
