Sum of this geometric sequence doesn't make sense

AI Thread Summary
The discussion centers on the sum of a geometric series defined by the expression Ʃ 2(4/3)^n from n=1 to 14. The original poster calculated the sum using an incorrect formula, leading to a result of approximately 330.74, while the answer sheet indicated around 441. The correct formula for the sum starting at n=1 is different from that starting at n=0, which caused confusion. The correct evaluation of the series yields a result close to 441, confirming the answer sheet's value. Understanding the correct application of the geometric series formula is crucial for accurate calculations.
It_Angel
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Homework Statement



14
Ʃ 2(4/3)^n
n=1

Homework Equations



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

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

It_Angel said:
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?
I believe the equation is working. a represents the first term in the series. In this case, what is a?
 
Yeah you got the typo.

Why is a not 2, as per tn=a*r^n?
 
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}
 
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.
 
It_Angel said:

Homework Statement



14
Ʃ 2(4/3)^n
n=1

Homework Equations



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



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?

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},\\<br /> 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
 

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