# series with geometric terms, find the sums.

• Nov 30th 2008, 11:09 AM
Legendsn3verdie
series with geometric terms, find the sums.
here is what i did. i guessu gotta know the formula (a)/(1-r) to find the sum. here is what i did no idea if i m right. Wondering if i could get some help on how this is done thanks. to my understand a is the first value of the series and r is related to the n.. not sure.

http://i176.photobucket.com/albums/w...titled-124.jpg
• Nov 30th 2008, 11:19 AM
Chris L T521
Quote:

Originally Posted by Legendsn3verdie
here is what i did. i guessu gotta know the formula (a)/(1-r) to find the sum. here is what i did no idea if i m right. Wondering if i could get some help on how this is done thanks. to my understand a is the first value of the series and r is related to the n.. not sure.

http://i176.photobucket.com/albums/w...titled-124.jpg

The series $\sum_{n=0}^{\infty}(-1)^n\frac{5}{4^n}$ is the same as $\sum_{n=0}^{\infty}5\left(-\frac{1}{4}\right)^n$, which has the form of a geometric series $\sum_{n=0}^{\infty}ar^n$

There are two things to note.

When you expand the series and it starts off like $a+ar+ar^2+\dots$, then the series sums up to $\boxed{\frac{a}{1-r}}$.

However, if you expand the series, and it starts off like $ar+ar^2+ar^3+\dots$, then the series sums up to $\frac{a}{1-r}-a=\boxed{\frac{ar}{1-r}}$.

In our case, the first term is $a=5$. Thus, the series $\sum_{n=0}^{\infty}5\left(-\frac{1}{4}\right)^n=\frac{5}{1+\frac{1}{4}}=\colo r{red}\boxed{4}$

You're answer was correct. Does this make sense?
• Nov 30th 2008, 11:32 AM
Legendsn3verdie
kind of but how do u know which way it expands

this series would expand like this right?:

(5) + (1/4) + (5/16) .......

how do i know if that is ar + ar^2 or a + ar ??
• Nov 30th 2008, 12:14 PM
Chris L T521
Quote:

Originally Posted by Legendsn3verdie
kind of but how do u know which way it expands

this series would expand like this right?:

(5) + (1/4) + (5/16) ....... Almost....take into account that there will be alternating signs.

how do i know if that is ar + ar^2 or a + ar ??

First, you identify a and r from the geometric series.

Since our series was $\sum_{n=0}^{\infty}5\left(-\tfrac{1}{4}\right)^n\sim\sum_{n=0}^{\infty}ar^n$, we see that ${\color{red}a=5}$ and ${\color{blue}r=-\tfrac{1}{4}}$

In expanding this series, we have $\underbrace{{\color{red}5}}_{\color{red}a}-\tfrac{5}{4}+\tfrac{5}{16}-\tfrac{5}{64}+\dots$. The first term in our series is $a$, thus we say it converges to $\frac{a}{1-r}\implies\frac{5}{1+\frac{1}{4}}=\color{red}\boxe d{4}$

However, what if our series was $\sum_{n=1}^{\infty}5\left(-\tfrac{1}{4}\right)^n$? We see that its similar to $\sum_{n=1}^{\infty}ar^n$, and can conclude [again] that ${\color{red}a=5}$ and ${\color{blue}r=-\tfrac{1}{4}}$.

This time, though, we see that it expands to $\underbrace{{\color{blue}-}\tfrac{{\color{red}5}}{{\color{blue}4}}}_{{\color {red}a}{\color{blue}r}}+\tfrac{5}{16}-\tfrac{5}{64}+\dots$. The first term in our series is $ar$, thus it converges to $\frac{a}{1-r}-a=\frac{ar}{1-r}\implies\frac{5\left(-\tfrac{1}{4}\right)}{1+\frac{1}{4}}=\color{red}\bo xed{-1}$

Does this clarify things??