Is there a shortcut for calculating summations such as this? 
Possible Duplicate:
Computing $\sum_{i=1}^{n}i^{k}(n+1-i)$ 

I'm curious in knowing if there's an easier way for calculating a summation such as
$\sum_{r=1}^nr(n+1-r)$
I know the summation $\sum_{r=1}^xr(n+1-r)$ is going to be a cubic equation, which I should be able to calculate by taking the first few values to calculate all the coefficients.  Then I can plug in $n$ for $x$ to get the value I'm looking for.  But it seems like there must be an easier way.
In the meantime, I'll be calculating this the hard way.
 A: If you already know the summations for consecutive integers and consecutive squares, you can do it like this:
$$\begin{align*}
\sum_{r=1}^n r(n+1-r)&=\sum_{r=1}^nr(n+1)-\sum_{r=1}^nr^2\\
&=(n+1)\sum_{r=1}^nr-\sum_{r=1}^nr^2\\
&=(n+1)\frac{n(n+1)}2-\frac{n(n+1)(2n+1)}6\\
&=\frac16n(n+1)\Big(3(n+1)-(2n+1)\Big)\\
&=\frac16n(n+1)(n+2)\;.
\end{align*}$$
Added: Which is $\dbinom{n+2}3$, an observation that suggests another way of arriving at the result. First, $r$ is the number of ways to pick one number from the set $\{0,\dots,r-1\}$, and $n+1-r$ is the number of ways to pick one number from the set $\{r+1,r+2,\dots,n+1\}$. Suppose that I pick three numbers from the set $\{0,\dots,n+1\}$; the middle number of the three cannot be $0$ or $n+1$, so it must be one of the numbers $1,\dots,n$. Call it $r$. The smallest number must be from the set $\{0,\dots,r-1\}$, and the largest must be from the set $\{r+1,r+2,\dots,n+1\}$, so there are $r(n+1-r)$ three-element subsets of $\{0,\dots,n+1\}$ having $r$ as middle number. Thus, the total number of three-element subsets of $\{0,\dots,n+1\}$ is $$\sum_{r=1}^nr(n+1-r)\;.$$ But $\{0,\dots,n+1\}$ has $n+2$ elements, so it has $\dbinom{n+2}3$ three-element subsets, and it follows that 
$$\sum_{r=1}^nr(n+1-r)=\binom{n+2}3=\frac{n(n+1)(n+2)}6\;.$$
A: Yes, linearity and a few memorized formulas:
$$\begin{array}{c l} \sum_{r=1}^n r(n+1-r) &=(n+1)\left(\sum_{r=1}^n r\right)-\sum_{r=1}^n r^2 \\ & = (n+1)\frac{n(n+1)}{2}-\frac{n(n+1)(2n+1)}{6} \\ & = \frac{n(n+1)(n+2)}{6}.\end{array}$$
A: Note that
$$
r(n+1-r)=n\binom{r}{1}-2\binom{r}{2}\tag{1}
$$
Using the identity
$$
\sum_r\binom{n-r}{a}\binom{r}{b}=\binom{n+1}{a+b+1}\tag{2}
$$
with $a=0$, we can sum $(2)$ for $r$ from $1$ to $n$ and get
$$
n\binom{n+1}{2}-2\binom{n+1}{3}\tag{3}
$$
Formula $(3)$ can be manipulated into more convenient forms, e.g.
$$
\left((n+2)\binom{n+1}{2}-2\binom{n+1}{2}\right)-2\binom{n+1}{3}\\[18pt]
3\binom{n+2}{3}-\left(2\binom{n+1}{2}+2\binom{n+1}{3}\right)\\[18pt]
3\binom{n+2}{3}-2\binom{n+2}{3}
$$
$$
\binom{n+2}{3}\tag{4}
$$
