<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] This is a possible way: 
$$\frac{9^{n+1}+4^{n+1}}{9^n+4^n}=6$$
Now: $$3^{2(n+1)}+2^{2(n+1)}=3\cdot 2\cdot 2^{2n}+2\cdot 3\cdot 3^{2n}$$ 
I obtain: $$3^{2(n+1)}+2^{2(n+1)}=3\cdot2^{2n+1}+2\cdot3^{2n+1}$$ 
In other words: $$3^{2n+1}(3-2)=2^{2n+1}(3-2)$$
The solution is $n=-\frac{1}{2}$.
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