Simplifying $\displaystyle\sum_{k=0}^{20}(k+4)\binom{23-k}{3}$ In trying to simplify my answer to a problem posted recently, I am trying to show that
$\displaystyle\sum_{k=0}^{20}(k+4)\binom{23-k}{3}=8\binom{24}{4}$.
I know that $\displaystyle\sum_{k=0}^{20}\binom{23-k}{3}=\binom{24}{4}$, but how can I simplify $\displaystyle\sum_{k=0}^{20}k\binom{23-k}{3}$?

Maple gives
$$\sum_{k=1}^n k\binom{m+n-k}{m}
  =\frac{(m+n)(m+n+1)}{(m+1)(m+2)}\binom{m+n-1}{m}\ ,$$
which can be checked by induction if you wish.  So your sum becomes
$$4\binom{24}{4}+\frac{23\times24}{4\times5}\binom{22}{3}
  =4\binom{24}{4}
    +\frac{23\times24}{4\times5}\frac{4\times20}{23\times24}\binom{24}{4}
  =8\binom{24}{4}\ .$$
