Double summation of the series of $\mathrm{C}_{n}^{m}(m+n+2)\left(\frac{1}{2}\right)^{m+n}$

Question

How to solve this double series problem:

$$\sum_{m=0}^{\infty}\sum_{n=0}^{m}\left[\mathrm{C}_{n}^{m}(m+n+2)\left(\frac{1}{2}\right)^{m+n}\right]$$

where $\mathrm{C}_{n}^{m}\equiv\frac{m!}{(m-n)!n!}$ is "combination without repetition".

The answer is 24.

Note. This double summation is midway in solving a "toss coin" question: What is the expected value of times until the sequence HH appears? For example, "THTHH" takes 5 times, "HH" takes 2 times, and "HTHH" takes 4 times. We can have the answer alternatively by simulation. (H is head, and T is tail)

I can only solve it by simulation.

Answer 1

Breaking the summation into three different summations, you get $\sum_{m=0}^\infty \frac{m}{2^m} \sum_{n=0}^m \frac{\binom mn}{2^n} + \sum_{m=0}^\infty \frac{1}{2^m}\sum_{n=0}^m \frac{n\binom mn}{2^n} + \sum_{m=0}^\infty \frac{2}{2^m} \sum_{n=0}^m \frac{\binom mn}{2^n}$. The first summation results in $\sum_{m=0}^\infty m\frac{3^m}{4^m}$ and the last summation results in $\sum_{m=0}^\infty 2\frac{3^m}{4^m}$. The second summation needs a bit of work. We can write $\binom mn$ as $\frac{m}{n} \binom{m-1}{n-1}$. Hence it becomes $\sum_{m=0}^\infty m\frac{1}{2^{m+1}} \sum_{n=1}^m \binom{m-1}{n-1}\frac{1}{2^{m-1}}$ which results in $\sum_{m=0}^\infty\frac{m}{4}\frac{3^{m-1}}{4^{m-1}}$. These three summations are A.G.P's which can be calucalted using standard techniques.