summation - Prove that $1^3 + 2^3 + ... + n^3 = (1+ 2 + ... + n)^2$

This is what I've been able to do:

Base case: $n = 1$

$L.H.S: 1^3 = 1$

$R.H.S: (1)^2 = 1$

Therefore it's true for $n = 1$.

I.H.: Assume that, for some $k \in \Bbb N$, $1^3 + 2^3 + ... + k^3 = (1 + 2 +...+ k)^2$.

Want to show that $1^3 + 2^3 + ... + (k+1)^3 = (1 + 2 +...+ (k+1))^2$

$1^3 + 2^3 + ... + (k+1)^3$

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

$ = (1+2+...+k)^2 + (k+1)^3$ by I.H.

Annnnd I'm stuck. Not sure how to proceed from here on.

Answer

HINT: You want that last expression to turn out to be $\big(1+2+\ldots+k+(k+1)\big)^2$, so you want $(k+1)^3$ to be equal to the difference

$$\big(1+2+\ldots+k+(k+1)\big)^2-(1+2+\ldots+k)^2\;.$$

That’s a difference of two squares, so you can factor it as

$$(k+1)\Big(2(1+2+\ldots+k)+(k+1)\Big)\;.\tag{1}$$

To show that $(1)$ is just a fancy way of writing $(k+1)^3$, you need to show that

$$2(1+2+\ldots+k)+(k+1)=(k+1)^2\;.$$

Do you know a simpler expression for $1+2+\ldots+k$?

(Once you get the computational details worked out, you can arrange them more neatly than this; I wrote this specifically to suggest a way to proceed from where you got stuck.)