The sum of the cubes of 11 to nn is equal to the square of the sum of 11 to nn

Theorem: For all natural numbers n≥1n \ge 1:

∑i=1ni3=(∑i=1ni)2\sum_{i=1}^n{i^3} = \left(\sum_{i=1}^n{i}\right)^2

Proof: We are going to prove this theorem using induction.

Let n=1n=1. Then:

∑i=1ni3=13=1=(1)2=(∑i=1ni)2\sum_{i=1}^n{i^3} = 1^3 = 1 = \left(1\right)^2 = \left(\sum_{i=1}^n{i}\right)^2

Let k∈Nk \in \mathbb{N} such that k≥1k \ge 1 and:

∑i=1ki3=(∑i=1ki)2\sum_{i=1}^k{i^3} = \left(\sum_{i=1}^k{i}\right)^2

Then for k+1k+1:

∑i=1k+1i3=∑i=1ki3+(k+1)3=(∑i=1ki)2+(k+1)(k+1)2=(∑i=1ki)2+k(k+1)2+1⋅(k+1)2=(∑i=1ki)2+2⋅k(k+1)2⋅(k+1)+(k+1)2=(∑i=1ki)2+2⋅(∑i=1ki)⋅(k+1)+(k+1)2=((∑i=1ki)+(k+1))2=(∑i=1k+1i)2\begin{split} \sum_{i=1}^{k+1}{i^3} &= \sum_{i=1}^k{i^3} + \left(k+1\right)^3 \\ &= \left(\sum_{i=1}^k{i}\right)^2 + \left(k+1\right) \left(k+1\right)^2 \\ &= \left(\sum_{i=1}^k{i}\right)^2 + k\left(k+1\right)^2 + 1\cdot\left(k+1\right)^2 \\ &= \left(\sum_{i=1}^k{i}\right)^2 + 2 \cdot \frac{k\left(k+1\right)}{2} \cdot \left(k+1\right) + \left(k+1\right)^2 \\ &= \left(\sum_{i=1}^k{i}\right)^2 + 2 \cdot \left(\sum_{i=1}^k{i}\right) \cdot \left(k+1\right) + \left(k+1\right)^2 \\ &= \left(\left(\sum_{i=1}^k{i}\right) + \left(k+1\right)\right)^2 \\ &= \left(\sum_{i=1}^{k+1}{i}\right)^2 \end{split}

Since:

using the principle of induction, the theorem is proven for all n≥1n \ge 1.