<s>[INST] Summing n times binomial(n,k) I'm trying to do $\sum_{n=a}^b \left( \begin{array}{rl} n \\ a \end{array} \right) n $ . 
Is there a formula, that anybody knows?
 [/INST] Use generating functions. First:
$\begin{align}
\sum_{n \ge 0} \binom{n}{a} z^n
  &= \frac{z^a}{(1 - z)^{a + 1}} \\
z \frac{\mathrm{d}}{\mathrm{d} z} \sum_{n \ge 0} \binom{n}{a} z^n
  &= \sum_{n \ge 0} n \binom{n}{a} z^n \\
  &= \frac{z^a (z + a)}{(1 - z)^{a + 2}}
\end{align}$
But also, to get partial sums you divide by $1 - z$:
$$
\sum_{k \ge 0} \left( \sum_{0 \le n \le k} n \binom{n}{k}\right) z^k
  = \frac{z^a (z + a)}{(1 - z)^{a + 3}}
$$
Now you are interested in the coefficient of $z^b$:
$\begin{align}
\sum_{0 \le n \le b} n \binom{n}{a}
  &= [z^b] \frac{z^a (z + a)}{(1 - z)^{a + 3}} \\
  &= [z^b] (z^{a + 1} + a z^a) \sum_{k \ge 0} (-1)^k \binom{-a - 3}{k} z^k \\
  &= [z^b] (z^{a + 1} + a z^a)
             \sum_{k \ge 0} \binom{a + 3 + k - 1}{a + 3 - 1} z^k \\
  &= [z^{b - a - 1}] \sum_{k \ge 0} \binom{a + 3 + k - 1}{a + 3 - 1} z^k
       + a [z^{b - a}]\sum_{k \ge 0} \binom{a + 3 + k - 1}{a + 3 - 1} z^k \\
  &= \binom{a + 2 + b - a - 1}{a + 2} + a \binom{a + 2 + b - a}{a + 2} \\
  &= \binom{b + 1}{a + 2} + a \binom{b + 2}{a + 2}
\end{align}$
</s>