summation of $\sum^n_{k=0} (n-k)^2$ I'm trying to find the recurrence of 
$$ T(n) = T (n-1) + n^2$$
After following the steps, 
$$T (n) = T (n-1) + n^2 = T (n-2) + (n-1)^2 + n^2  $$
$$T (n) = T (n-2) + (n-1)^2 + n^2 = T(n-3) + (n-3)^2 + (n-1)^2 + n^2   $$
$$T (n) = T (n-3) + (n-3)^2 + (n-1)^2 + n^2 =T (n-4) + (n-4)^2 + (n-3)^2 + (n-1)^2 + n^2 $$
i generalize recurrence relation at the kth step of the recursion, which is 
$$ T(n) = T (n-k) + \sum^n_{k=0} (n-k)^2$$
Just wondering, 
What is the summation of 
$$\sum^n_{k=0} (n-k)^2$$
is it the same as $$ n(n+1)(2n+1)/6$$
?

Hint: It is the same as $\sum_{k=0}^n k^2$.
