If you find the sum of the cubes of the first n natural numbers, which n gives you a sum of 4356?
10
11
12
13
Explanation:
Sum of cubes of first n natural numbers = $[\dfrac{n(n + 1)}{2}]^{2}$
Plug in each of the options to find the answer which is = 4356.
n = 10; Sum of cubes of first 10 numbers = $[\dfrac{10 × 11}{2}]^{2}$ = 3025.
n = 11; Sum of cubes of first 11 numbers = $[\dfrac{11 × 12}{2}]^{2}$ = 4356.