Sum of series involving binomial coefficients Evaluate the following;
$$\sum_{r=0}^{50} (r+1) ^{1000-r}C_{50-r}$$
Using $^{n}C_{r}=^{n}C_{n-r}$ we get $\sum_{r=0}^{50} (r+1) ^{1000-r}C_{950}$
but I am not getting how to solve $\sum_{r=0}^{50} r \cdot \hspace{0.5 mm} ^{1000-r}C_{950}$
 A: Set $50-r=u$
$$\sum_{u=0}^{50}(51-u)\binom{950+u}{950}=\sum_{u=0}^{50}\{1002-(951+u)\}\binom{950+u}{950}$$
$$=1002\sum_{u=0}^{50}\binom{950+u}{950}-951\sum_{u=0}^{50}\binom{951+u}{951}$$
Now $\displaystyle\sum_{u=0}^{50}\binom{950+u}{950}$  is the coefficient of $x^{950}$ in $$\displaystyle\sum_{u=0}^{50}(1+x)^{950+u}$$
A: This is a hockey stick made of hockey sticks. Expand each term $(r+1)\binom{1000-r}{950}$ into a column of $r+1$ copies of $\binom{1000-r}{950}$, then add up the rows using the hockey stick identity, then add up the rows sums using the hockey stick identity.
$$\begin{array}{rrrcrl}
\displaystyle\binom{1000}{950}&+\displaystyle2\binom{999}{950}&+\displaystyle3\binom{999}{950}&\dots&+\displaystyle50\binom{950}{950}\\\hline\\
=\displaystyle\binom{1000}{950}&+\displaystyle\binom{999}{950}&+\displaystyle\binom{999}{950}&\dots&+\displaystyle\binom{950}{950} &\stackrel{\text{H.S.}}=\displaystyle\binom{1001}{951}
\\\\ & +\displaystyle\binom{999}{950}&+\displaystyle\binom{998}{950}&\dots &+\displaystyle\binom{950}{950}
&\stackrel{\text{H.S.}}=+\displaystyle\binom{1000}{951}
\\\\ &&+\displaystyle\binom{998}{950}&\dots &+\displaystyle\binom{950}{950}
&\stackrel{\text{H.S.}}=+\displaystyle\binom{999}{951}
\\\\&&&\ddots&\vdots\;\;\;\;&\;\;\;\;\;\;\;\;\;\;\;\vdots
\\\\&&&&+\displaystyle\binom{950}{950}&\stackrel{\text{H.S.}}=+\displaystyle\binom{951}{951}
\\\\
&&&&&\stackrel{\text{H.S.}}=\displaystyle\boxed{\binom{1002}{952}}
\end{array}$$
