How to sum:$\sum_{k=1}^{n}(k²-k)$ without using this: $\sum_{k=1}^{n}(k²)=\frac{n(n+1)(2n+1)}{6}$? I have tried to find the sum of this series  $\sum_{k=1}^{n}(k^2-k)$ without 
using  $\sum_{k=1}^{n}(k^2)=\frac{n(n+1)(2n+1)}{6}$ but i can't  .
Can anyone explain this to me if there is a way to do it?
Note : Thank you for your help  
 A: $$\sum_{k=1}^n k^2-k=\frac{1}{3}\sum_{k=1}^n 3k(k-1)=\frac{1}{3}\sum_{k=1}^n (k+1-(k-2))k(k-1)$$
$$=\frac{1}{3}\left(\sum_{k=1}^n (k+1)k(k-1)-(k-2)k(k-1)\right)$$
$$=\frac{1}{3}\left(0-0+3.2.1-0+4.3.2-3.2.1+\cdots +(n+1)n(n-1)-n(n-1)(n-2)\right)$$
$$=\boxed{\dfrac{1}{3}(n+1)n(n-1)}$$
A: $$\sum_{k=1}^n k(k-1) = \frac{d^2}{dx^2}\sum_{k=0}^n x^k  \text{ at } x = 1$$
Now evaluate $$\frac{d^2}{dx^2}\frac{x^{n+1}-1}{x-1} $$
A: Using basic facts from finite calculus (as explained, for example, in Section $2.6$ of Graham, Knuth, & Patashnik, Concrete Mathematics), you can write
$$\sum_{k=1}^n(k^2-k)=\sum_{k=1}^nk^{\underline 2}=\left[\frac{k^{\underline 3}}3\right]_1^{n+1}=\frac{(n+1)n(n-1)}3=\frac{n(n^2-1)}3\;.$$
A: $$\sum_{k=1}^n(k^2-k)=2\sum_{k=1}^n{k\choose 2}=2\sum_{k=1}^n\left[{k+1\choose 3}-{k\choose 3}\right]=2{n+1\choose 3}$$
A: Here's one approach:
Since one is adding a sum of quadratic expressions, heuristically one expects the formula for the sum to be cubic in $n$. So, we can form the ansatz
$$\sum_{k = 1}^n (k^2 - k) = A n^3 + B n^2 + C n + D.$$
(We can also replace $k^2 - k$ with $k^2$, in which case we recover the familiar formula mentioned in the original question.)
A cubic polynomial is determined by its value at four points, so we need only substitute four convenient values and solve the resulting system in $A, B, C, D$. (Hint Evaluating both sides at $n = 0$ gives $D = 0$.)
With these values in hand, we can make the argument rigorous simply by proving our cubic formula using induction.
