Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$

Another way (by Euler, I think), from the geometric sum:

$$1 + x + x^2 + \cdots + x^n = \frac{x^{n+1}-1}{x-1} \tag{1}$$

Differentiate both sides and multiply by $x$:

$$x + 2 x^2 + 3 x^3 + \cdots + n x^{n} = \frac{n x^{n+2}-(n+1) x^{n+1} +x}{(x-1)^2} \tag{2}$$

Differentiating 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 compute the derivative on the RHS in $(2)$ (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

$$\begin{align} g(x)&=\frac{x^{n+1}-1}{x-1}\\ &=\frac{\left(1+(x-1)\right)^{n+1}-1}{x-1}\\ &={n+1 \choose 1}+{n+1 \choose 2}(x-1)+{n+1 \choose 3}(x-1)^2+O\left((x-1)^3\right) \tag{4} \end{align}$$

Deriving, multiplying by $x$ and deriving again: $$(g'(x) \, x)'={n+1 \choose 2}+{n+1 \choose 3}2 \, x + O(x-1) \tag{5}$$ 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:
$$ \sum_{k=1}^n k^3={n+1 \choose 2}+{n+1 \choose 3}6+{n+1 \choose 4}6$$

I like this visual proof, due to Man-Keung Siu. It appeared in the March 1984 issue of Mathematics Magazine.

See also two more proofs (as well as this one) in Roger Nelson's Proofs Without Words: Exercises in Visual Thinking.

You can easily prove it by induction.

One way to find the coefficients, assuming we already know that it's a degree $3$ polynomial, is to calculate the sum for $n=0,1,2,3$. This gives us four values of a degree $3$ polynomial, and so we can find it.

The better way to approach it, though, is through the identity $$ \sum_{t=0}^n \binom{t}{k} = \binom{n+1}{k+1}. $$ This identity is true since in order to choose a $(k+1)$-subset of $n+1$, you first choose an element $t+1$, and then a $k$-subset of $t$.

We therefore know that $$ \sum_{t=0}^n A + Bt + C\binom{t}{2} = A(n+1) + B\binom{n+1}{2} + C\binom{n+1}{3}. $$ Now choosing $A=0,B=1,C=2$, we have $$ A+Bt + C\binom{t}{2} = t^2. $$ Therefore the sum is equal to $$ \binom{n+1}{2} + 2\binom{n+1}{3}. $$

Notice that $(k+1)^3 - k^3 = 3k^2 + 3k + 1$ and hence

$$(n+1)^3 = \sum_{k=0}^n \left[ (k+1)^3 - k^3\right] = 3\sum_{k=0}^n k^2 + 3\sum_{k=0}^n k + \sum_{k=0}^n 1$$

which gives you

$$\begin{align} \sum_{k=1}^n k^2 & = \frac{1}{3}(n+1)^3 - \frac{1}{2}n(n+1) - \frac{1}{3}(n+1) \\ & = \frac{1}{6}(n+1) \left[ 2(n+1)^2 - 3n - 2\right] \\ & = \frac{1}{6}(n+1)(2n^2 +n) \\ & = \frac{1}{6}n(n+1)(2n+1) \end{align}$$

Proof (by induction)

Basis: Check it for n = 1 (it works out).

Induction: Assume the result is true for a given value of $n$. That is, assume $$ \sum_{k = 1}^n k^2 = \frac{n(n+1)(2n+1)}{6}. $$ Try to show that the result holds for $n+1$. $$ \begin{align*} \sum_{k = 1}^{n+1} k^2 &= (n+1)^2 + \sum_{k=1}^n k^2\\ &= (n+1)^2 + \frac{n(n+1)(2n+1)}{6}\\ &= \frac{6(n+1)^2 + n(n+1)(2n+1)}{6}\\ &= \frac{(n+1)(n+1+1)(2(n+1)+1)}{6}. \end{align*} $$