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posted by Anonymous

posted 161d 3h ago

We have,

$\frac{1}{(5+x)^2} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} n x^{n-1}}{5^{n+1}}$

when

. We want to find

$\sum_{n=1}^{\infty}\frac{n3^{n}}{5^{n}}.$

So,

$\sum_{n=1}^{\infty}\frac{n3^{n}}{5^{n}} = \sum_{n=1}^{\infty}\frac{n((-1)(-1)3)^{n}}{5^{n}}$

$\sum_{n=1}^{\infty}\frac{n((-1)(-3))^{n}}{5^{n}}$

$\sum_{n=1}^{\infty}\frac{n(-1)^{n}(-3)^{n}}{5^{n}}$

$\sum_{n=1}^{\infty}\frac{n(-1)^{n+1}(-3)^{n-1}(-1)^{-1}(-3)}{5^{n+1}5^{-1}}$

$\left[\sum_{n=1}^{\infty}\frac{n(-1)^{n+1}(-3)^{n-1}}{5^{n+1}}\right](3 \cdot 5)$

$\left[\frac{1}{(5+(-3))^2}\right](3 \cdot 5)$

$\frac{15}{4}$

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