I need to find a closed formula for $\sum_{i=1}^{n}i^2$. Of course, I can prove this by induction, and I know that the required formula is $\frac{n(n+1)(2n+1)}{6}$, but I don't know how to prove it combinatorially.
I found that $\frac{n(n+1)(2n+1)}{6}={n+1 \choose 3}+{n+2 \choose 3}$, but I cannot exceed from here. Please help me.
Summing the first $n$ squares can be geometrically interpreted as the sum of the areas of $k$ by $k$ squares for $1\leq k\leq n$.
If we arrange these $k$ by $k$ blocks in ascending order, placing a $k+1$ by $k+1$ block to the right of a $k$ by $k$ block, creating in essence a staircase, we may make the following observation:
The first row of this staircase has area $T_n$, where $T_n$ represents the $n$-th triangle number, i.e., $T_n=\binom{n+1}{2}$. The second row of this staircase has area $T_n-T_1$. In general, the $k$-th row of this staircase has area $T_n-T_{k-1}$.
This observation allows us to say that $\sum_{k=1}^nk^2=\sum_{k=1}^n(T_n-T_{k-1})$.
Expanding upon this, we have: $$\begin{align} \sum_{k=1}^nk^2 &=\sum_{k=1}^n(T_n-T_{k-1})\\ &=nT_n-\sum_{k=1}^nT_{k-1}\\ &=nT_n-\sum_{k=1}^n\frac{k(k-1)}{2}\\ &=nT_n-\frac{1}{2}\sum_{k=1}^nk^2+\frac{1}{2}\sum_{k=1}^nk\\ &=nT_n-\frac{1}{2}\sum_{k=1}^nk^2+\frac{1}{2}T_n \end{align}$$
Solving for $\sum_{k=1}^nk^2$, we see that $$\sum_{k=1}^nk^2=\frac{2}{3}T_n\left(n+\frac{1}{2}\right)=\frac{2}{3}\frac{n(n+1)}{2}\frac{(2n+1)}{2}=\frac{n(n+1)(2n+1)}{6}$$
Note that this method suggest a formula for $\sum_{k=1}^nk^{m+1}$ in terms of $\sum_{k=1}^nk^{m}$ for all naturals $n,m$.
