<s>[INST] How can I simplify $\frac{3^{n+1} + 3^{n-1}}{3^{n+2} + 3^n}$? This problem got me thinking hard: 
$$\frac{3^{n+1} + 3^{n-1}}{3^{n+2} + 3^n}$$
Considering the fact that the common base is '$3$', I decided to do this:
$$\log_3\left(\frac{(n+1)(n-1)}{(n+2)(n)}\right)$$
$$\log_3\left(\frac{n^2-n+n+1}{n^2+2n}\right)$$
$$\log_3\left(\frac{n^2 + 1}{n^2 + 2n}\right)$$
$$\log_3\left(\frac{1}{2n}\right)$$
I am not 100% sure if have simplified the problem. I need some clarification.
 [/INST] $$\frac{3^{n+1}+3^{n-1}}{3^{n+2}+3^n}=\frac{3^{n-1}(3^2+1)}{3^{n-1}(3^3+3)}=\frac{10}{30}=\frac 13$$
This shows your simplification is incorrect for a number of reasons:


*

*$\log(3^{n+1}+3^{n-1})\neq \log(3^{n+1})\log(3^{n-1})$. The $\log$ rule works the other way: $\log(a)+\log(b)=\log(ab)$

*You cannot simplify $\frac{n^2+1}{n^2+2n}$ to $\frac{1}{2n}$: big no-no. You can simplify multiplied expressions, not $summed$ expressions. For instance $\frac{n(n+1)}{(n+3)(n+1)}=\frac{n}{n+3}$ is fine.

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