summation - Determine infinite sum with shifting binomial coefficients

I am looking for a way to find analytically the following sum

$\sum_{i=0}^\infty \frac{1}{6^i}
\begin{pmatrix}
i+k\\ i
\end{pmatrix}
= \Big(\frac65\Big)^{1+k}$,

for some integer $k>0$. The answer I displayed here was found by Wolfram Alpha. I looked at partial sums, but it involves hypergeometric functions. Is there any way around this? Many thanks in advance.

Answer

Note that $$F\left(k+1\right)=1+\sum_{i\geq1}\dbinom{i+k+1}{i}\frac{1}{6^{i}}
$$ $$=1+\sum_{i\geq1}\left(\dbinom{i+k}{i}+\dbinom{i+k}{i-1}\right)\frac{1}{6^{i}}
$$ $$=1+\sum_{i\geq1}\dbinom{i+k}{i}\frac{1}{6^{i}}+\sum_{i\geq1}\dbinom{i+k}{i-1}\frac{1}{6^{i}}
$$ $$=\sum_{i\geq0}\dbinom{i+k}{i}\frac{1}{6^{i}}+\frac{1}{6}\sum_{i\geq0}\dbinom{i+k+1}{i}\frac{1}{6^{i}}
$$ $$=F\left(k\right)+\frac{F\left(k+1\right)}{6}
$$ hence $$F\left(k+1\right)=\frac{6}{5}F\left(k\right)
$$ and since $$F\left(1\right)=\sum_{i\geq0}\dbinom{i+1}{i}\frac{1}{6^{i}}=\sum_{i\geq0}\frac{i+1}{6^{i}}=\left(\frac{6}{5}\right)^{2}
$$ we have $$F\left(k+1\right)=\left(\frac{6}{5}\right)^{k+2}.
$$