# Evaluating $\sum\limits_{n=0}^{20} \frac{(-1)^{n}2^{n+1}}{3^{n}},$

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.

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Are you only trying to compute a partial sum? You don't want $\sum_{n=1}^{\infty} (-1)^n 2^{n+1}/3^n$? – Dimitrije Kostic Nov 15 '11 at 0:44
The index in the sum should start at n=0. Are you aware of the geometric sum? Read here en.wikipedia.org/wiki/Geometric_progression#Geometric_series – leonbloy Nov 15 '11 at 0:45
Yes, just the partial sum. I know it's wierd, but my professor put it on our test and I had no idea how to do it then. Now he offered test corrections because the average was below 60 and I still don't know how to do it. – Caleb Jares Nov 15 '11 at 0:45
$1+a+a^2+\cdots +a^n=\frac{1-a^{n+1}}{1-a}$ if $a \ne 1$. You can verify this by multiplying left side by $1-a$. – André Nicolas Nov 15 '11 at 0:50
Thanks! That worked. – Caleb Jares Nov 15 '11 at 1:07

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 $$S = a\frac{r^n - 1}{r - 1}.$$
$$\sum_{n=1}^{20}\frac{(-1)^n2^{n+1}}{3^n}=2\sum_{n=1}^{20}\left(-\frac{2}{3}\right)^n=2\left(-\frac{2}{3}\right)\frac{\left(-\frac{2}{3}\right)^{20}-1}{\left(-\frac{2}{3}-1\right)}=$$ $$=\left(-\frac{4}{3}\right)\left(-\frac{3}{5}\right)\frac{2^{20}-3^{20}}{3^{20}}=\frac{4}{5}\frac{2^{20}-3^{20}}{3^{20}}$$