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?

Yet another take. Start with the definition of falling factorial powers:
$\begin{align}
  x^{\underline{r}}
    = x (x - 1) \dotsm (x - r + 1)
\end{align}$
so that:
$\begin{align}
  \Delta n^{\underline{r}}
    &= (n + 1)^{\underline{r}} - n^{\underline{r}} \\
    &= r n^{\underline{r - 1}} \\
  \sum_{0 \le k < n} k^{\underline{r}}
    &= \frac{n^{\underline{r + 1}}}{r + 1}
\end{align}$
(the last one is also easy to prove by induction, or otherwise).

Now note that:
$\begin{align}
  n^2
    = n^{\underline{2}} + n^{\underline{1}}
\end{align}$
so we can write:
$\begin{align}
   \sum_{0 \le k \le n} k^2
     &= \sum_{0 \le k < n + 1}
          \left( k^{\underline{2}} + k^{\underline{1}} \right) \\
     &= \frac{(n + 1)^{\underline{3}}}{3}
          + \frac{(n + 1)^{\underline{2}}}{2} \\
     &= \frac{(n + 1) n (n - 1)}{3} + \frac{(n + 1) n}{2} \\
     &= \frac{(2 n + 1) (n + 1) n}{6}
\end{align}$
