In my research, I've come across the following weighted sum of binomial coefficients: $$\sum_{i=1}^k i\cdot{2k \choose k+i}.$$ According to Wolfram Alpha, this equals exactly the quantity $$\frac{1}{2}(k+1){2k \choose k+1}.$$ After trying to prove this, I've come up shorthanded. I have two questions.
- Is there a direct procedure (or a combinatorial argument) that yields this expression?
- Is there a general class of weighted sums that could be solved in a similar way?
I'll give a different solution from the two in the linked question.
Now to view it as computing $E|X-k|$ where $X \sim Bin\left(2k,\frac 12\right)$, it's quite natural to do the following:
$$\sum_{i=1}^k i\cdot{2k \choose k+i} = \sum_{i=1}^k (k+i-k) \binom{2k}{k+i} \\ = \sum_{i=1}^k (k+i) \binom{2k}{k+i} - k \sum_{i=1}^k \binom{2k}{k+i}\\ = 2k \sum_{i=1}^k \binom{2k-1}{k+i-1} - k \cdot \frac{2^{2k} - \binom{2k}{k}}{2}\\ = 2k \cdot \frac{2^{2k-1}}{2} - k\cdot \frac{2^{2k} - \binom{2k}{k}}{2}\\ = \frac{k}{2} \binom{2k}{k} = \frac{1}{2}(k+1){2k \choose k+1}. \blacksquare $$
Note the above quantity is actually equal to $2^{2k-1}E|X-k|$, not $E|X-k|$.