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Geometric Series Question

  • Thread starter brojas7
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  • #1
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[itex]\sum^{\infty}_{k=4}[/itex] [itex]\frac{1}{5^{k}}[/itex]


ar^n=a/(1-r)


Here is what I did:

a=5
r=1/625

(1/5)/(1-1/625)=(1/5)/(624/625)=625/3120=125/624

The Answer is 1/500

Where am I going wrong?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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[itex]\sum^{\infty}_{k=4}[/itex] [itex]\frac{1}{5^{k}}[/itex]


ar^n=a/(1-r)


Here is what I did:

a=5
r=1/625

(1/5)/(1-1/625)=(1/5)/(624/625)=625/3120=125/624

The Answer is 1/500

Where am I going wrong?
You've got a and r wrong. a is the first term in the series and r is the common ratio. Try again.
 
  • #3
Fredrik
Staff Emeritus
Science Advisor
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Can you rewrite the sum in the form
$$\frac{1}{5^4}\sum_{i=0}^\infty(\text{something})_i?$$ This isn't absolutely necessary, but if you do, you can use the version of the formula with a=1.
 
  • #4
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You've got a and r wrong. a is the first term in the series and r is the common ratio. Try again.
i meant a to be 1/5...
the first term was 1/625
 
  • #5
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Can you rewrite the sum in the form
$$\frac{1}{5^4}\sum_{i=0}^\infty(\text{something})_i?$$ This isn't absolutely necessary, but if you do, you can use the version of the formula with a=1.
If I do it that way,r would equal 1/5 right?

$$\frac{1}{5^4}\sum_{i=0}^\infty(\text{(1/5)^n})?$$
 
  • #6
Fredrik
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Yes, assuming that you meant what I hope you meant when you wrote ar^n=a/(1-r), which doesn't really make sense. :smile:
 
  • #7
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It worked, thank you so much
 
  • #8
The sum of an infinite geometric series of the form ##\displaystyle\sum_{n=0}^\infty ar^n=\dfrac{a}{1-r}##, for ##|r|<1##.

Your given series starts off with ##n=4##. Think you can rewrite the series to make it start with ##n=0##?
 
  • #9
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The sum of an infinite geometric series of the form ##\displaystyle\sum_{n=0}^\infty ar^n=\dfrac{a}{1-r}##, for ##|r|<1##.

Your given series starts off with ##n=4##. Think you can rewrite the series to make it start with ##n=0##?
I didn't know in order for the formula to work I had to make n=0. Thanks
 

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