by math induction on n,
for n = 1, the statement is true since
[tex]1^3=\left(\frac{1(1+1)}{2}\right)^2[/tex]
assume that the statement is true for n = k ie
[tex]1^3+2^3+...+k^3=\left(\frac{k(k+1)}{2}\right)^2[/tex]
we will use the equation for n = k to show that the statement is also true for n = k + 1.
observe
[tex]1^3+2^3+...+k^3+(k+1)^3
\\=\left(\frac{k(k+1)}{2}\right)^2+(k+1)^3
\\=\left(\frac{k^2}{4}+k+1\right)(k+1)^2
\\=\left(\frac{k^2+4k+4}{4}\right)(k+1)^2
\\=\left(\frac{k+2}{2}\right)^2(k+1)^2
\\=\left(\frac{(k+1)+1}{2}\right)^2(k+1)^2
\\=\left(\frac{(k+1)((k+1)+1)}{2}\right)^2[/tex]
Q.E.D.
