Explanation of a method to compute $\sum_{k \le n} k^2$ I was searching for methods to compute $\sum_{k\le n} k^2$. I stumbled across this (which is an answer provided by Gareth Rees to this question).

"..Represent $k^2$ in terms of falling powers (easy by inspection in this case, but you can use Stirling subset numbers to convert): $$ k^2 = k^{\underline 2} + k^{\underline 1}$$
Sums of falling powers are easy, just like integration of ordinary powers, except for the treatment of limits: $$ \sum_{k=1}^n k^{\underline 2} + k^{\underline 1} = \bigg({1\over 3}k^{\underline 3} + {1\over 2}k^{\underline 2}\bigg)\ \bigg|^{n+1}_0$$
And then convert back into ordinary powers (by expansion, or using signed Stirling cycle numbers): $$ {1\over 3}((n+1)^3 - 3(n+1)^2 + 2(n+1)) + {1\over 2}((n+1)^2 - (n+1))..."$$

Could someone explain what is going on in this answer and provide some useful and explanatory links?
Thank you.

Instead of falling powers I might use Binomial Coefficients:
$$ k^2 = 2\, \frac{k(k-1)}{2} + k = 2\binom{k}{2}  + \binom{k}{1}$$
Now if we add over all $k$ we can get other binomial coefficients:
$$ \sum_{k \leq n} \binom{k}{2} = \binom{n+1}{3} \text{ and } \sum_{k \leq n} \binom{k}{1} = \binom{n+1}{2}$$
This feels just like the forumla $\int x^k = \tfrac{1}{k+1} x^{k+1} + C$ and finally 
$$ \sum_{k \leq n} k^2 
= 2\sum_{k \leq n} \binom{n}{2}  + \sum_{k \leq n} \binom{n}{1}
= 2\binom{n+1}{3} + \binom{n+1}{2}
= 2\cdot  \frac{(n+1)n(n-1)}{6} + \frac{(n+1)n}{2}$$
This recovers the correct formula since:
$$ 2\cdot  \frac{(n+1)n(n-1)}{6} + \frac{(n+1)n}{2} = \frac{n(n+1)(2n+1)}{6}$$
