$n^2 + (n-1)^2 + \dots +2^2 + 1^2$ equals what? $$n^2 + (n-1)^2 + \dots +2^2 + 1^2 = \text{??}$$
I am reading that this equals 
$$\frac{1}{3} n \left( n + \frac{1}{2} \right) (n+1) $$
But have no clue how.. The thing that strikes me most is the fact that the latter has a cubed factor when expanded (i.e. in big O notation it's O$(n^3)$). 
 A: This can be done in several ways (and has been done many times on this site). Let's
try something a bit different.
Let
$$f(x)=\frac13x(x+\frac12)(x+1).$$
Then by differentiating we get
$$
g(x)=f'(x)=x^2+x+\frac16.
$$
It is easy to verify that
$$
g(x)-g(x-1)=2x.
$$
Integrating this gives us
$$
f(x)-f(x-1)=x^2+C
$$
for some constant $C$. As $f(0)=f(-1)=0$, we see that $C=0$, so for all $n$ we have
$$
f(n)-f(n-1)=n^2
$$
Thus we are at the point Andrea Mori described. The rest is easy
$$
1^2+2^2+\cdots+n^2=(f(1)-f(0))+(f(2)-f(1))+\cdots+(f(n)-f(n-1))=f(n)-f(0)=f(n).
$$
A: For a geometric "proof" or explanation with respect to how to make sense of the resulting identity for the sum of n consecutive squares, see the entry here.
It uses the equivalent identity: $$1^2 + 2^2 + \cdots + (n - 1)^2 + n^2 = \sum_{k = 1}^n k^2 = \frac 13(n)\left(n + \frac 12\right)(n+1)  =\dfrac {n(n+1)(2n+1)}{6} $$ which can easily be "proven" by induction on $n$.
A: Let
$$
S(n)=1^2+2^2+\cdots+n^2.
$$
Since $S(n)-S(n-1)=n^2$ one can suspect that $S(n)$ is a cubic polynomial in $n$. Obviously, $S(n)\in\Bbb Z$ for all $n=1$, $2$, $3\dots$ and one knows that the ring of numeric polynomals, i.e. the polynomials in $\Bbb Q[X]$ which take integer values on $\Bbb N$ is freely generated over $\Bbb Z$ by the binomial coefficients
$$
\binom Xk=\frac1{k!}X(X-1)\cdots(X-k+1).
$$
Thus one may attempt to write
$$
S(n)=a\binom X0+b\binom n1+c\binom n2+d\binom n3,
$$
for some $a,b,c,d\in\Bbb Z$ (since $\binom Xk$ has degree $k$, those with $k\geq4$ will not enter). Some fiddling allows to find the right coefficients and confirm the formula.
A: Hint: Use induction on $n$
$$n=1:\quad1^2=\frac{1}{3}(1)\left(\frac{3}{2}\right)(2)=1$$
Assume statement is true for $n$, then
\begin{align*}
(n+1)^2+n^2+\dots+2^2+1^2&=n^2+2n+1+\frac{1}{3}(n)\left(n+\frac{1}{2}\right)(n+1)\\
\end{align*}
Then expand and factorise and you'll get your answer.
