Znajdź sumę szeregu potęgowego

[Klefix]

\(\sum\limits^\infty _{n=1} \frac{x^n3^n}{n+1}\)

[maxkor]

\(x\sum_1^\infty {{x^n}\over {n+1}}=\sum_1^\infty {{x^{n+1}}\over {n+1}}=\sum_2^\infty {{x^n}\over n}=-\ln(1-x)-x \implies \sum_1^\infty {{x^n}\over {n+1}}={{-\ln(1-x)}\over x}-1 \implies\sum_1^\infty {{(3x)^n}\over {n+1}}={{-\ln(1-3x)}\over {3x}}-1\)