Believe it or Not

[BigGlenntheHeavy]

For any number of terms, it is true that:

(1+2+3+...+n)^2 = 1^3+2^3+3^3+...+n^3.

 

[galactus]

I believe so. As we know, the sum of the first n integers is \(\displaystyle \frac{n(n+1)}{2}\)

The sum of the first n cubes is given by \(\displaystyle \frac{n^{2}(n+1)^{2}}{4}\)

The latter is the square of the former.