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I need to find the sum of the following series as an explicit formula: $ \sum_{n=1}^\infty n(n+1)(\frac{1}{2})^n $. I can't express it as a geometric series $ \sum_{n=1}^\infty ar^{n-1} = \frac{a}{1-r} $ since the term n(n+1) is not constant.

I also considered that this series might be a Taylor expansion of a function $$ \sum_{n=1}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n $$ but I don't see how $ \frac{f^{(n)}}{n!} $ Could give me a constant term n(n+1). Does anyone have an idea of how I could proceed? Thanks!

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Let me try. We have $$\sum_{n=0}^\infty x^n = \frac{1}{1-x}.$$

Taking derivative two times, we have

$$\sum_{n=2}^\infty n(n-1)x^{n-2} = \frac{2}{(1-x)^3}.$$

So, we have $$\sum_{n=2}^\infty n(n-1)x^{n-1} = \frac{2x}{(1-x)^3},$$

or $$\sum_{n=1}^\infty n(n+1)x^{n} = \frac{2x}{(1-x)^3}.$$

Substituting $x=\frac{1}{2}$, you get the result.

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Hint: What is the Maclaurin series of $f(x) = \frac{1}{1-x}$ if $|x| < 1$? What does this tell you about the Maclaurin series of $f'(x)$, $f''(x)$, etc.?

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