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

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.

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

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

Let's look for a pattern

1³ = 1 = 1²
1³ + 2³ = 9 = 3²
1³ + 2³ + 3³ = 36 = 6²
1³ + 2³ + 3³ + 4³ = 100 = 10²
1³ + 2³ + 3³ + 4³ + 5³ = 225 = 15²

See the pattern yet?
1 + 2 = 3
3 + 3 = 6
6 + 4 = 10
10 + 5 = 15

First we add 2, then we add 3, then 4, then 5, etc

So, continuing the pattern, we get:
1³ + 2³ + 3³ + . . . + 6³ = 21²
1³ + 2³ + 3³ + . . . + 7³ = 28²
1³ + 2³ + 3³ + . . . + 8³ = 36²
1³ + 2³ + 3³ + . . . + 9³ = 45²
1³ + 2³ + 3³ + . . . + 10³ = 55²

Answer: C

Cheers,
Brent