Calculate the sum of cubes from 1 to100

it can be proven that $\displaystyle\sum_{k=1}^{n} k^{3} = (\displaystyle\sum_{k=1}^{n} k )^{2}$
Since $\displaystyle\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$
$\displaystyle\sum_{k=1}^{n} k^{3} = (\frac{n(n+1)}{2})^{2} = \frac{n^{2} (n+1)^{2}}{4}$
In this case $n=100$ so $\displaystyle\sum_{k=1}^{n} k^{3} = \frac{10000(10201)}{4} = 25502500$

                                        Very easily calculated as  (100* 101/2)^2 = 5050^2 = 25502500.