# Sum of this geometric sequence doesn't make sense!

1. Sep 24, 2012

### It_Angel

1. The problem statement, all variables and given/known data

14
Ʃ 2(4/3)^n
n=1

2. Relevant equations

Sn=a(1-r^n)/(1-r)

3. The attempt at a solution

2(1-[4^14]/[3^14])/(-1/3)=330.74

However, the answer sheet gives ~441 as the answer, and I confirmed it by doing it by hand. Why is the equation not working? What's wrong?

Last edited: Sep 24, 2012
2. Sep 24, 2012

### Ninty64

I'm assuming that second 2 is a typo and should be an n.
$\sum ^{14}_{n=1} 2(\frac{4}{3})^n$

I believe the equation is working. $a$ represents the first term in the series. In this case, what is $a$?

3. Sep 24, 2012

### It_Angel

Yeah you got the typo.

Why is a not 2, as per tn=a*r^n?

4. Sep 24, 2012

### Ninty64

The sum of a geometric series is defined as:
$a+ar+ar^2+ar^3+...+ar^{n-1} = a\frac{1-r^n}{1-r}$

If n started at 0, then a would be 2.
Since n starts at 1, in order to form a geometric series we must group it as following:
$\frac{8}{3} + \frac{8}{3}(\frac{4}{3}) + \frac{8}{3}(\frac{4}{3})^2 + ... + \frac{8}{3}(\frac{4}{3})^{13}$

5. Sep 24, 2012

### Mentallic

It_Angel, you might not have put this together for yourself so I'll just mention it.

The reason the geometric sum

$$a+ar+ar^2+...+ar^n = a\frac{1-r^{n+1}}{1-r}$$

Is because we can simply factor out an "a" on the left side, and then if we compare both sides,

$$a(1+r+r^2+...+r^n)=a\left(\frac{1-r^{n+1}}{1-r}\right)$$

Clearly we can just divide both sides by "a" to get what the geometric sum (starting from 1) is equal to.

Anyway, the moral of the story is if you can't figure out what a should be, all you need to do is factor out some value such that the geometric sum inside the factor begins at 1, and then you know the value you factored out must be a. Or even more easily: Whatever the first value of the sum is, that is equal to a.

6. Sep 24, 2012

### Ray Vickson

Two problems: (i) incorrect evaluation of result; and (ii) incorrect formula. We have
$$a \sum_{n=0}^N r^n = a \frac{1-r^{N+1}}{1-r},\\ a \sum_{n=1}^N r^n = a \frac{r - r^{N+1}}{1-r}.$$
The formula starting at n = 1 is a bit different from that starting at n = 0.

Anyway, I get $2 \sum_{n=0}^{14} (4/3)^n \doteq 442.9854833,$ while $2 \sum_{n=1}^{14} (4/3)^n \doteq 440.9854833.$ I cannot get your 330.74 from either formula.

RGV