User:Beneficii/mathdraft

Draft:

It can also be proven using mathematical induction and the arithmetic series:

We are proving for an integer n:

For $$n \ge 1$$, $$\left ( \sum_{k=1}^n k \right )^2 = \sum_{k=1}^n k^3$$

We look first at the base case:

For $$n = 1$$, $$\left ( \sum_{k=1}^1 k \right )^2 = \sum_{k=1}^1 k^3$$

$$1^2 = 1^3$$

$$1 = 1$$

Thus it is proven for the base case.

Now, for the inductive step. Assume, that it is true for a case $$n = x$$:

$$\left ( \sum_{k=1}^x k \right )^2 = \sum_{k=1}^x k^3$$

For $$n = x + 1$$, $$\begin{align} \left ( \sum_{k=1}^{x+1} k \right )^2 &= \left ( \sum_{k=1}^x k \right )^2 + \left ( \sum_{k=1}^{x+1} k \right )^2 - \left ( \sum_{k=1}^x k \right )^2 \\ &= \left ( \sum_{k=1}^x k \right )^2 + \left ( \frac{(x+1)(x+2)}{2} \right )^2 - \left ( \frac{x(x+1)}{2} \right )^2 \\ &= \left ( \sum_{k=1}^x k \right )^2 + \left ( \frac{(x+1)^2(x+2)^2}{4} \right ) - \left ( \frac{x^2(x+1)^2}{4} \right ) \\ &= \left ( \sum_{k=1}^x k \right )^2 + \frac{(x+1)^2(x+2)^2 - x^2(x+1)^2}{4} \\ &= \left ( \sum_{k=1}^x k \right )^2 + \frac{(x+1)^2[(x+2)^2 - x^2]}{4} \\ &= \left ( \sum_{k=1}^x k \right )^2 + \frac{(x+1)^2[(x^2 + 4x + 4 - x^2]}{4} \\ &= \left ( \sum_{k=1}^x k \right )^2  + \frac{(x+1)^2(4x + 4)}{4} \\ &= \left ( \sum_{k=1}^x k \right )^2  + \frac{4(x+1)^2(x + 1)}{4} \\ &= \left ( \sum_{k=1}^x k \right )^2  + (x+1)^2(x + 1) \\ &= \left ( \sum_{k=1}^x k \right )^2  + (x+1)^3 \\ &= \sum_{k=1}^x k^3  + (x+1)^3 \\ &= \sum_{k=1}^{x+1} k^3 \end{align}$$

We have proved it for the inductive step.