Infinite Sum with Geometric and Combinatorial terms

Question

I was working on a probability question which at one point brought me to the sum:

$$\sum_{k=0}^{\infty} \Big{(}\frac{4}{6}\Big{)}^k {4+k \choose k}$$

I put it in Wolfram Alpha and obtained the answer $243$. That answer also gives me the correct answer to my original probability problem.

I have tried re expressing the combinatorial term as factorials, but cannot seem to simplify it any further. I could have sworn this type of sum was covered in my undergraduate studies...but cannot find any references.

Any help would be appreciated.

Answer 1

In general you have for $|x|<1$ $$\frac 1{(1-x)^{n+1}} = \sum_{k=0}^\infty\binom{n+k}k x^k$$

In your case you get as sum $3^5 = 243$.