That's some series you got there

The $n^{\text{th}}$ term of a series $1+\frac{1}{6}+\frac{1}{72}+...$ is given by $a(\frac{1}{4})^n+b(\frac{1}{3})^n$ . Find the sum to infinity of the series.

Where, $a$ and $b$ are integers

NB: The answer will be in the form of $\frac{c}{d}$ . Express it as $10c+d$