Find $n$ if $\frac{9^{n+1}+4^{n+1}}{9^n+4^n} = 6$ Find $n$ if $$\frac{9^{n+1}+4^{n+1}}{9^n+4^n} = 6$$

In this video they show a shortcut and say $n=-1/2$ without any explanation. 
Key observation here is that the geometric mean of $9$ and $4$ is $6$.
It seems numerator and denominator are partial sums of geometric series, but I don't know how to proceed. Any help?

$ \frac{9^{n+1} + 4^{n+1}}{9^n + 4^n} = 6 $
$ 9^{n+1} + 4^{n+1} = 6\cdot 9^n + 6 \cdot 4^n $
$3 \cdot 9^n = 2 \cdot 4^n $
$ 3^{2n+1} = 2^{2n+1}$
$(\frac{3}{2})^{2n+1} = 1$
So, $2n+1 = 0 $
$ n = -\frac{1}{2}$
