The sum of the cubes of the first n positive integers has the formula [n(n + 1)/2]^2. Thus the sum of the cubes of the first 10 positive integers is [10(11)/2]^2 = [(2 x 5 x 11)/2]^2 = 55^2.VJesus12 wrote:What is the sum of the cubes of the first ten positive integers?
A. 10^3
B. 45^2
C. 55^2
D. 100^2
E. 100^3
[spoiler]OA=C[/spoiler]
Source: Manhattan GMAT
Alternative solution:
Another way to reiterate the above fact is the sum of cubes of the first n positive integers is the square of the sum of the first n positive integers. Thus the sum of the cubes of the first 10 positive integers is
(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)^2 = 55^2
(Notice that adding the first 10 positive integers is much faster than actually adding the cubes of the first 10 integers.)
Answer: C
