Question: I want to calculate

$$\sum\_{k=0}^\infty\binom{k+3}k(0.2)^k$$

to get the exact value of it. I have excel and other tools to help me so it is fine if it is computationally expensive. Is there a clear and repeatable way to solve this infinite series? Thank you. This is my first post and be sure to give me some suggestions as well.

Answer: 
It’s a general fact that

$$\sum\_{k\ge 0}\binom{k+n}nx^k=\frac1{(1-x)^{n+1}}\;.$$

You can prove this by induction on $n$, starting with the geometric series

$$\frac1{1-x}=\sum\_{k\ge 0}x^k$$

and differentiating repeatedly with respect to $x$. You want the case $n=3$:

$$\sum\_{k\ge 0}\binom{k+3}kx^k=\sum\_{k\ge 0}\binom{k+3}3x^k=\frac1{(1-x)^4}\,.$$

Now just substitute $x=0.2$.
Using the [negative binomial coefficient](https://math.stackexchange.com/a/217647/13854) $\binom{-4}{k}$, we have
$$
\begin{align}
\sum\_{k=0}^\infty\binom{k+3}{k}(0.2)^k
&=\sum\_{k=0}^\infty\binom{-4}k(-1)^k(0.2)^k\\
&=(1-0.2)^{-4}\\[6pt]
&=\frac{625}{256}
\end{align}
$$