Prove that $(1+2+3+\cdots + n)^2 = 1^3+2^3+3^3+\cdots + n^3$ for every $n \in \mathbb{N}$

Question

Prove that $(1+2+3+\cdots + n)^2 = 1^3+2^3+3^3+\cdots + n^3$ for every $n \in \mathbb{N}$.

Proof. We will use mathematical induction. If $n = 1$, then we have $(1)^2= 1^3 = 1$. We must show that $S_n$ implies $S_{n+1}$. Assume that for $n \in \mathbb{N}$, the statement $(1+2+3+\cdots + n)^2 = 1^3+2^3+3^3+\cdots + n^3$ holds. Then

$\begin{align} &\left[\displaystyle\sum_{n \in \mathbb{N}}^n n\right]^2+(n+1)^3 \\ =& \displaystyle\sum_{n \in \mathbb{N}}^n n^3+(n+1)^3 \\ =& \displaystyle\sum_{n \in \mathbb{N}}^n n^3 + n^3 + 3n^2 +3n + 1 \end{align}$

How do I finish carrying out the induction? I can't seem to manipulate the right hand side properly get the form $\left[\displaystyle\sum_{n \in \mathbb{N}}^n n+1\right]^2$.

Answer 1

Consider these:

• $$\sum_{i=1}^ni=\frac{n(n+1)}2$$
• $$\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6$$
• $$\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4$$

Answer 2

Both sides are polynomials of degree $4$. Both sides take the values $0,1,9,36,100$ for $n=0,1,2,3,4$. So they are the same polynomial!

Answer 3

Interesting graphical proof with applet in The Sum of the Cubes of the First N Natural Numbers.