<s>[INST] 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?
 [/INST] Start by dividing top and bottom by $4^n$:
$$\frac{\frac{9^{n+1}}{4^n} + \frac{4^{n+1}}{4^n}}{\frac{9^n}{4^n} + 1} = 6.$$
Substitute $u = \frac{9^n}{4^n}$ to get
$$\frac{9u + 4}{u + 1} = 6 \iff 9u + 4 = 6(u + 1) \iff u = \frac{2}{3}.$$
Therefore,
$$\left(\frac{9}{4}\right)^n = \frac{2}{3} \implies n = -\frac{1}{2}.$$
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