Evaluate:$\frac{1}{2^{101}}\sum_{k=1}^{51} \sum_{r=0}^{k-1}\binom{51}{k}\binom{50}{r}$

Question

Evaluate:$$\frac{1}{2^{101}}\sum_{k=1}^{51} \sum_{r=0}^{k-1}\binom{51}{k}\binom{50}{r}$$ My Attempt:

I did try writing the series $(1+x)^{50}$and $(1+x)^{50}$ separately,then multiplied but could not determine the power of $x$ whose coefficient should be found.Can Vandermonde's Identity be put to use here.

Answer 1

Using the identity $$ \binom{51}{k}=\binom{50}{k}+\binom{50}{k-1}, $$ we get $$ \sum_{k=1}^{51} \sum_{r=0}^{k-1}\binom{51}{k}\binom{50}{r}=\sum_{k=1}^{51} \sum_{r=0}^{k-1}\binom{50}{k}\binom{50}{r}+\sum_{k=1}^{51} \sum_{r=0}^{k-1}\binom{50}{k-1}\binom{50}{r}.$$ Observe that the former is equal to $$ \sum_{k=0}^{50} \sum_{r=0}^{k-1}\binom{50}{k}\binom{50}{r}= \sum_{ 0\leq r<k\leq 50}\binom{50}{k}\binom{50}{r} = \sum_{ 0\leq k<r\leq 50}\binom{50}{k}\binom{50}{r}=\sum_{k=0}^{50} \sum_{r=k+1}^{50}\binom{50}{k}\binom{50}{r}. $$ Also note that the latter is equal to $$\begin{eqnarray} \sum_{k=1}^{51} \sum_{r=0}^{k-1}\binom{50}{k-1}\binom{50}{r}=\sum_{k=0}^{50}\sum_{r=0}^{k}\binom{50}{k}\binom{50}{r}. \end{eqnarray}$$ Gathering them together yields $$ \sum_{k=0}^{50} \sum_{r=k+1}^{50}\binom{50}{k}\binom{50}{r}+\sum_{k=0}^{50}\sum_{r=0}^{k}\binom{50}{k}\binom{50}{r}=\sum_{k=0}^{50} \sum_{r=0}^{50}\binom{50}{k}\binom{50}{r} = 2^{50}\cdot2^{50} = 2^{100}. $$ Hence the answer is $\frac{1}{2}$.

Answer 2

Here's an alternative solution.

Observe \begin{align} S=&\ \sum^{51}_{k=1}\sum^{k-1}_{r=0}\binom{51}{k}\binom{50}{r} =\ \sum^{51}_{k=1}\sum^{k-1}_{r=0}\binom{51}{51-k}\binom{50}{50-r}\\ =&\ \sum^{50}_{k'=0} \sum^{50}_{r'=k'}\binom{51}{k'}\binom{50}{r'} = \sum^{51}_{k'=0} \sum^{50}_{r'=k'}\binom{51}{k'}\binom{50}{r'}. \end{align} Hence it follows \begin{align} 2S= \sum^{51}_{k=0}\sum_{r=0}^{50}\binom{51}{k}\binom{50}{r}= \sum^{51}_{k=0}\binom{51}{k}\sum_{r=0}^{50}\binom{50}{r}=2^{51}\cdot 2^{50} \end{align} which means \begin{align} S = 2^{100}. \end{align} Hence you arrive at $S/2^{101} = 1/2$.