Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 3 down vote favorite
1

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.

share|improve this question
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
3  
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
4  
$1+a+a^2+\cdots +a^n=\frac{1-a^{n+1}}{1-a}$ 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
show 2 more comments

2 Answers

up vote 3 down vote accepted

Define $$S = a + ar + \ldots + a{r^{n - 1}}$$ for $r \ne 1$, and multiply by $r$ to get $$rS = ar + a{r^2} + \ldots + a{r^n}.$$ Subtracting $S$ from $rS$ gives $$rS - S = a{r^n} - a$$ or $$S(r - 1) = a{r^n} - a$$ so $$S = a\frac{{{r^n} - 1}}{{r - 1}}.$$

share|improve this answer
You can also multiply $S$ by $(-1)/(-1)$ to get $S = a(1 - {r^n})/(r - 1)$, which is used in some books. Moreover, note that if $|r| < 1$ $S \to a/(r - 1)$ as $n \to \infty $ otherwise it diverges. – glebovg Oct 22 '12 at 4:58
up vote 1 down vote

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

share|improve this answer
I'm disinclined to upvote this because it contains no words. Most importantly, there's no text to highlight the use of the geometric series. – Douglas S. Stones Oct 20 '12 at 6:08
The 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. – DonAntonio Oct 20 '12 at 17:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.