Calculate $\sum_{i=0}^\infty i\binom{i+10}{i}(\frac{1}{2})^i$

Question

I'm not great with summation and its various techniques, so go easy on me. If someone can point me in the right direction I would be very grateful.

This is the sum in display mode: $$\sum_{i=0}^\infty i\binom{i+10}{i}\frac{1}{2^i}$$

By using the symmetry rule for binomial coefficients I end up with this: $$\frac{1}{10!}\sum_{i=0}^\infty \frac{i}{2^i}\prod_{k=1}^{10}(k+i)$$ which doesn't seem to help. I tried working something else on the binomial to get rid of i and I got: $$11\sum_{i=0}^\infty \frac{1}{2^i}\binom{i+10}{i-1}$$ but yet again, I'm stuck.
I have a feeling that I have to somehow simplify the sum and then use the recursive method?

Answer 1

Hint: A well-known series (see here, e.g.) is:

$$\frac{1}{(1-x)^{11}} = \sum_{k=0}^\infty \binom{10+k}{k}x^k$$

Its derivative is close to what you want:

$$\frac{d}{dx} \frac{1}{(1-x)^{11}} = \sum_{k=1}^\infty k\binom{10+k}{k}x^{k-1}$$ At the end evaluate at $x=\frac12$ of course.

Answer 2

Hint:

You may note that $$ \sum_{i\ge0}i\binom{10+i}i x^i= x\frac{d}{dx}\sum_{i\ge0}\binom{10+i}i x^i, $$ reducing the problem to known series.

$$=x\frac{d}{dx}\sum_{i\ge0}\binom{-11}i (-x)^i=x\frac{d}{dx}\frac1{(1-x)^{11}}=\left.\frac{11x}{(1-x)^{12}}\right|_{x=\frac12}=22528.$$