Evaluate the series $\sum\limits_{x=0}^{\infty} \binom{4+x}{x} \frac{1}{2^x}$ I am trying to find the value of
$$\sum\limits_{x=0}^{\infty} \binom{4+x}{x} \frac{1}{2^x}$$
and using the formula for $\binom{n}{r}$, I have simplified it to
$$\sum\limits_{x=0}^{\infty} \frac{(x+4)(x+3)(x+2)(x+1)}{24\cdot 2^x}$$
but I have no idea how to proceed. It's question 9 from this link:
https://www.nanomath.org/sample.pdf

The problem is: Evaluate $\sum\limits_{x=0}^{\infty} \binom{4+x}{x} \frac{1}{2^x}$
Let $$S(n) = \sum\limits_{x=0}^{\infty} \binom{n+x}{x} \frac{1}{2^x}$$
So we have $$S(4) = \sum\limits_{x=0}^{\infty} \binom{4+x}{x} \frac{1}{2^x}$$
$$S(3) = \sum\limits_{x=0}^{\infty} \binom{3+x}{x} \frac{1}{2^x}$$
$$S(4) - S(3) = \sum\limits_{x=0}^{\infty} \left(\binom{4+x}{x} - \binom{3+x}{x}\right) \frac{1}{2^x}$$
$$ = \frac{1}{2}\sum\limits_{x=1}^{\infty} \binom{3+x}{x-1} \frac{1}{2^{x-1
}}$$
$$ = \frac{1}{2}\sum\limits_{x=0}^{\infty} \binom{4+x}{x} \frac{1}{2^x} =\frac{S(4)}{2}$$ by change of variable.
So $S(4) = 2S(3) = 4S(2) = 8S(1) = 16S(0)$ where
$$S(0) = \sum\limits_{x=0}^{\infty} \binom{0+x}{x} \frac{1}{2^x} = 2$$
So $$S(4) = 32$$
