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?

Another way (by Euler, I think), starting from the geometric sum:
$$1 + x + x^2 + \cdots + x^n = \frac{x^{n+1}-1}{x-1} \tag{1}$$
Differentiate both sides:
$$1 + 2 x + 3 x^2 + \cdots + n x^{n-1} = \frac{n x^{n+1}-(n+1) x^{n} +1}{(x-1)^2} \tag{2}$$
Multiply by $x$ and differentiate once more. We get on the LHS
$$1 + 2^2 x + 3^2 x^2 + \cdots + n^2 x^{n-1} \tag{3}$$
which, evaluated at $x=1$ gives our desired sum $\sum_{k=1}^n k^2$. Hence, we just need to multiply by $x$ the RHS of $(2)$, compute the  (eg, with L'Hopital rule) and evaluate it at $x \to 1$.
It should be evident that this procedure also can be applied (though it also turns more cumbersome) for sums of higher powers.

(Update) here's a concrete computation, applying the binomial theorem on the RHS of $(1)$ and doing a series expansion around $x=1$. Let
$$\begin{align}
g(x)&=\frac{x^{n+1}-1}{x-1}\\
&=\frac{\left(1+(x-1)\right)^{n+1}-1}{x-1}\\
&={n+1 \choose 1}+{n+1 \choose 2}(x-1)+{n+1 \choose 3}(x-1)^2+O\left((x-1)^3\right) \tag{4}
\end{align}$$
Differentiating, multiplying by $x$ and differentiating again we get $$(g'(x) \, x)'={n+1 \choose 2}+{n+1 \choose 3}2 \, x + O(x-1) \tag{5}$$
... which evaluated at $x=1$ gives the desired answer:
$${n+1 \choose 2}+2{n+1 \choose 3} =\frac{n(n+1)(2n+1)}{6}  $$
We can apply the same procedure for higher powers. For example:
$$ \sum_{k=1}^n k^3={n+1 \choose 2}+{n+1 \choose 3}6+{n+1 \choose 4}6$$
