I know the sum of the series $2 - \frac{4}{3} + \frac{8}{9} - \cdots + \frac{(-1)^{20}2^{21}}{3^{20}}$ is equal to $\sum\limits_{n=0}^{20} \frac{(-1)^{n}2^{n+1}}{3^{n}},$ but I don't know how to calculate the sum without manually entering it into the calculator.
Evaluating $\sum\limits_{n=0}^{20} \frac{(-1)^{n}2^{n+1}}{3^{n}},$
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$\begingroup$
calculus
sequences-and-series
summation
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0@gary Bad idea. – 2012-07-12
2 Answers
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Define $S = a + ar + \cdots + ar^{n - 1}$ for $r \ne 1$, and multiply by $r$ to get $rS = ar + ar^2 + \cdots + ar^n.$ Subtracting $S$ from $rS$ gives $rS - S = ar^n - a$ or $S(r - 1) = ar^n - a.$ Therefore $S = a\frac{r^n - 1}{r - 1}.$
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$\sum_{n=1}^{20}\frac{(-1)^n2^{n+1}}{3^n}=2\sum_{n=1}^{20}\left(-\frac{2}{3}\right)^n=2\left(-\frac{2}{3}\right)\frac{\left(-\frac{2}{3}\right)^{20}-1}{\left(-\frac{2}{3}-1\right)}=$ $=\left(-\frac{4}{3}\right)\left(-\frac{3}{5}\right)\frac{2^{20}-3^{20}}{3^{20}}=\frac{4}{5}\frac{2^{20}-3^{20}}{3^{20}}$
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0The OP's *apparent* level (by his question) makes words reduntant in this easy, basic and straightforward case. People shouldn't be used to get all chewed up and digested, imo. – 2012-10-20