Sum of cubes proof [duplicate]

Solution 1:

By induction:

for $n=1$ it works. Then, suppose it works for a $n$. Then, $$(1+...+n+(n+1))^2=\underbrace{(1+...+n)^2}_{=1^3+...+n^3\ by\ hyp.}+2(1+...+n)(n+1)+(n+1)^2$$$$=1^3+...+n^3+(n+1)\big(2\underbrace{(1+...+n)}_{=\frac{n(n+1)}{2}}+(n+1)\big)$$$$=1^3+...+n^3+(n+1)\big(n(n+1)+(n+1)\big)$$$$=1^3+...+n^3+(n+1)(n+1)(n+1)$$$$=1^3+...+n^3+(n+1)^3.$$

Q.E.D.

Solution 2:

Here an illustration. The surface is the square of the sum. Prove that each added layer (in different color) is a the wanted cube and you will be done.

Solution 3:

This solution assumes you are allowed to use $$ V_1 = \sum_{k=1}^{n} k = \frac{n(n+1)}{2}\\ V_2 = \sum_{k=1}^{n}k^2 = \frac{n(n+1)(2n+1)}{6} $$ Use the perturbation method (set the sum of cubes equal to $V_3$. Consider $$ S_n = \sum_{k=1}^{n}k^4 $$ then $$ S_n + (n+1)^4 = 1 + \sum_{k=1}^{n}(k+1)^4 = 1+S_n + 4 \sum_{k=1}^{n} k^3 + 4 \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k +n $$ Obviously $S_n$ cancels out, you know $V_1$ and $V_2$, so you can get the value for $V_3$ and see that it's equal to $\frac{(n(n+1))^2}{4}$.