<s>[INST] Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$ I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:
$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$
I really have no idea why this statement is true. Can someone please explain why this is true and if possible show how to arrive at one given the other?
 [/INST] Another take, similar to Euler's (?) proof given by @leonbloy. We know that if:
$\begin{align}
   A(z)
     &= \sum_{n \ge 0} a_n z^n
\end{align}$
then (writing $\mathtt{D}$ for the derivative):
$\begin{align}
  z \mathtt{D} A(z)
    &= \sum_{n \ge 0} n a_n z^n
\end{align}$
Also:
$\begin{align}
  \frac{A(z)}{1 - z}
    &= \sum_{n \ge 0} \left( \sum_{0 \le k \le n} a_n \right) z^n \\
  \frac{1}{1 - z}
    &= \sum_{n \ge 0} z^n
\end{align}$
This you can repeat and combine. In our case, we get that:
$\begin{align}
   \sum_{n \ge 0} n^2
     &= (z \mathtt{D})^2 \frac{1}{1 - z} \\
     &= \frac{z + z^2}{(1 - z)^3} \\
   \sum_{n \ge 0} \left( \sum_{0 \le k \le n} k^2 \right) z^n
     &= \frac{z + z^2}{(1 - z)^4}
\end{align}$
We are interested in the coefficient of $z^n$:
$\begin{align}
  [z^n] \frac{z + z^2}{(1 - z)^4}
    &= [z^n] \frac{z}{(1 - z)^4} + [z^n] \frac{z^2}{(1 - z)^4} \\
    &= [z^{n - 1}] (1 - z)^{-4} + [z^{n - 2}] (1 - z)^{-4} \\
    &= (-1)^{n - 1} \binom{-4}{n - 1} + (-1)^{n - 2} \binom{-4}{n - 2} \\
    &= \binom{n - 1 + 4 - 1}{4 - 1} + \binom{n - 2 + 4 - 1}{4 - 1} \\
    &= \binom{n + 2}{3} + \binom{n + 1}{3} \\
    &= \frac{(n + 2) (n + 1) n}{3!} + \frac{(n + 1) n (n - 1)}{3!} \\
    &= \frac{(2 n + 1) (n + 1) n}{6}
\end{align}$
This approach is less messy than the cited one (no horrible derivatives and then l'Hôpital thrice).
</s>