Proof without using induction [duplicate]

With linear algebra:

You can do it with linear algebra, by noticing that the map $\Sigma$ which maps a sequence $u_n$ to the sequence $u_0 + \dots + u_n$ has, as a left inverse, the map $\Delta$, mapping the sequence $u_n$ to the sequence $u_n - u_{n-1}$; since, for any sequence $u_n = f(n)$ where $f$ is a polynomial of degree $d$, $\Delta u$ is polynomial of degree $d-1$, we see that $\Sigma$ maps polynomials of degree $d$ to polynomials of degree $d+1$, and we can find which ones by linear algebra.

With combinatorics:

An even simpler way is to use the standard basis $e_d$ where $e_d$ is the sequence $e_{d,n} = \binom{n+d}{d}$, for then we have $$(\Delta e_d)_n = e_{d,n} - e_{d,n-1} = \binom{n+d}{d} - \binom{n+d-1}{d} = \binom{n+d-1}{d-1} = (e_{d-1})_{n}.$$ Then we only need to write $n^2$ as a linear combination of $e_{0,n} = 1$, $e_{1,n} = n$ and $e_{2,n} = n(n-1)/2$.

With integrals:

A third way is to use integrals: namely, with simple linear algebra we find that $$ n^2 = \int_{n-1}^{n} (x^2+x+1/6) dx,$$ so that $$ 0+1+\dots+n^2 = \int_0^{n} (x^2+x+1/6) dx = \frac{1}{6} (2n^3+3n^2+n).$$

With geometry:

A fourth way is the usual geometric proof, where you can tile together six pyramids in a parallelepiped.

Evaluate the telescoping sum

$$\sum_{k=1}^n k^3-(k-1)^3=n^3$$

But $k^3-(k-1)^3=3k^2-3k-1$. Thus,

$$\sum_{k=1}^n k^3-(k-1)^3=n^3=\sum_{k=1}^n (3k^2-3k-1)$$

Isolate the sum over $k^2$ and use the result of the sum of an arithmetic sum and you will have it!

This methodology, using a telescoping sum, works for higher powers. So, it can be used to find the sum of $k^m$ for any integer $m$.