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?

$$\dfrac{9(9^n) + 4(4^n)}{9^n + 4^n} = 6 $$ divide numerator and denominator by $9^n$:
$$ \dfrac{9 + 4\left(\frac{4}{9}\right)^n}{1 + \left(\frac{4}{9}\right)^n} = 6$$
Let $\left(\frac{4}{9}\right)^n = a$:
$$ \dfrac{9+ 4a}{1 + a} = 6 $$
$$a = \frac{3}{2} $$
$$ \left(\frac{4}{9}\right)^n =\left(\frac{2}{3}\right)^{2n} = \left(\frac{2}{3}\right)^{-1}$$
$$ n = -\frac{1}{2}$$
