The sum of the first n cubes is a perfect square for every n, since 1^3+...+n^3 = (n(n+1)/2 )^2. For which value of n is this sum equal to 3025?

Question

Accepted Answer

10. We need (n(n+1)/2 )^2 = 3025, so n(n+1)/2 = 55, giving n(n+1) = 110 = 10 x 11. Thus n = 10. Check: (10 × 11/2 )^2 = 55^2 = 3025.

The sum of the first $n$ cubes is a perfect square for every $n$ , since $1^{3} + ... + n^{3} = (\frac{n ( n + 1 )}{2})^{2}$ .
For which value of $n$ is this sum equal to 3025?