Prove $1^3+2^3+...+n^3 = (1+2+...+n)^2$ for all positive integers $n$.

Question

My approach is to solve this by induction.

Base case: $n=1$

$1^3 = 1^2 = 1$

Inductive Step: Suppose that $1^3+2^3+...+n^3 = (1+2+...+n)^2$ holds for all positive integers $n$. We use that to show that $1^3+2^3+...+n^3 + (n+1)^3 = (1+2+...+n+(n+1))^2$

$1^3+2^3+...+n^3 + (n+1)^3 = (1+2+...+n)^2 + (n+1)^3$

And here is where I can no longer proceed. I know there must be some trick/computation that I am not seeing that will make this proof easier. Help? Thanks!!

Answer 1

You can use the fact

$$(1+2+3+\ldots+n)=\frac{n(n+1)}{2}$$ $$\left(\frac{n(n+1)}{2}\right)^2=\frac{n^2(n+1)^2}{4}$$ Then you just have to prove that $$\frac{n^2(n-1)^2}{4}+(n+1)^3=\frac{(n+1)^2(n+2)^2}{4}$$

Answer 2

$$ \left(\sum_{i=1}^{n+1}i\right)^2 = \left(\sum_{i=1}^n i + (n+1)\right)^2 = \left(\sum_{i=1}^n i\right)^2 + 2(n+1)\sum_{i=1}^n i + (n+1)^2 $$ The second and third terms are $$ 2(n+1)\frac{n(n+1)}{2} + (n+1)^2 = (n+1)^2(n+1) = (n+1)^3 $$ Thus $$ \left(\sum_{i=1}^{n+1}i\right)^2 = \left(\sum_{i=1}^n i \right)^2 +(n+1)^3 $$

Answer 3

Hint: \begin{align*} (1+2+\dots+n+(n+1))^2&=(1+2+\dots+n)^2+2(1+2+\dots+n)(n+1)+(n+1)^2 \\ &=1^3+\dots+n^3+2(1+2+\dots+n)(n+1)+(n+1)^2\\ &=1^3+\dots+n^3+n(n+1)^2+(n+1)^2\\ &=1^3+\dots+n^3+(n+1)^3\end{align*}