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\displaystyle\begin{aligned} \sum_{n\geq 1} x^n = \frac{x}{1-x} \implies \sum_{n\geq 1} n x^{n-1} & = \frac{d}{dx} \frac{x}{1-x} \\ & = \frac{1}{(x-1)^2} \\ \implies \sum_{n\geq 1} 4n x^{n+1} & = \frac{4x^2}{(x-1)^2} \\ \implies \sum_{n\geq 0} \frac{4n}{3^{n+1}} & = \frac{4x^2}{(x-1)^2} \bigg|_{x=\frac{1}{3}} = 1
\displaystyle\begin{aligned} \sum_{n\geq 1} \frac{(-1)^n}{n^2} & = \sum_{n\geq 1} \frac{1}{(2n)^2} - \sum_{n\geq 1} \frac{1}{(2n-1)^2} \\ & = \frac{\pi^2}{24} - \left( \sum_{n\geq 1} \frac{1}{n^2} - \sum_{n\geq 1} \frac{1}{(2n)^2} \right) \\ & = -\frac{\pi^2}{12}
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$ k =\frac{a^n+b^n}{(ab)^{n-1}+1}\in\mathbb{N})\Rightarrow \exists c\in\mathbb{N}\; (k = c^n) $
$ k =\frac{a^n+b^n}{(ab)^{n-1}+1}\in\mathbb{N})\Rightarrow \exists c\in\mathbb{N}\; (k = c^n) $
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