Find a formula for $1+2^{3}+3^{3}+\cdots+n^{3}$.
Let's calculate the first few terms to get an idea:

$1 ; 1+8=9 ; 1+8+27=36 ; 1+8+27+64=100 ; 1+8+27+64+125=225 ; \ldots$

We immediately notice that all the terms are squares, the squares of the sequence $1,3,6,10,15$, which we recognize as the sequence $1,(1+2),(1+2+3)$, etc. We will try to prove the following formula by induction:

$$
1+2^{3}+3^{3}+\cdots+n^{3}=\left(\frac{n(n+1)}{2}\right)^{2}
$$

The base case is easy. Now suppose the formula is true for $n$ and calculate

$1+\cdots+n^{3}+(n+1)^{3}=\frac{n^{2}(n+1)^{2}}{4}+(n+1)^{3}=\frac{(n+1)^{2}\left(n^{2}+4(n+1)\right)}{4}=\frac{(n+1)^{2}(n+2)^{2}}{4}$

which completes the induction.