- #1

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1st and 3rd partial sums of the sequence An=(-2)^n+5

----I don't even know what formula to use to start this problem

and sum of hte geometric series:

2/3 - 4/9 + 8/27 - ...

I think I use this formula for this one: S=A1/1-r

Please help

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- Thread starter TonyC
- Start date

- #1

- 86

- 0

1st and 3rd partial sums of the sequence An=(-2)^n+5

----I don't even know what formula to use to start this problem

and sum of hte geometric series:

2/3 - 4/9 + 8/27 - ...

I think I use this formula for this one: S=A1/1-r

Please help

- #2

HallsofIvy

Science Advisor

Homework Helper

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A

Yes, the sum of an infinite sum a+ ar+ ar

Here, you have (2/3)+ (2/3)(-2/3)+ (2/3)(-2/3)

- #3

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For the second I have come up with an answer of .518

Am I correct?

- #4

VietDao29

Homework Helper

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And also, how did you come up with .518 in #2???

Your first term is 2 / 3. And all you need to do is to find r. So what do you get for r?

Viet Dao,

- #5

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For the SVietDao29 said:_{1}is correct, but I think you should re-check your answer for S_{3}, S_{3}= A_{1}+ A_{2}+ A_{3}.

And also, how did you come up with .518 in #2???

Your first term is 2 / 3. And all you need to do is to find r. So what do you get for r?

Viet Dao,

((-2)

#2:For r, I have r=1/3

- #6

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Can anyone lend some advice?

- #7

VietDao29

Homework Helper

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Nope, you are not doing it correctly.

S_{n} is the sum of **the first** n terms.

So S_{3} is the sum of the first **3 terms**. So:

S_{3} = A_{1} + A_{2} + A_{3} = ...

Note that they are not asking for A_{3}, they are asking for **S**_{3}.

So what do you get for S_{3}?

--------------------

How can you come up with r = 1 / 3???

[tex]a_1 = \frac{2}{3}[/tex]

[tex]a_2 = -\frac{4}{9} = a_1r[/tex]

So again, what is r?

Viet Dao,

S

So S

S

Note that they are not asking for A

So what do you get for S

--------------------

How can you come up with r = 1 / 3???

[tex]a_1 = \frac{2}{3}[/tex]

[tex]a_2 = -\frac{4}{9} = a_1r[/tex]

So again, what is r?

Viet Dao,

Last edited:

- #8

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I am still baffled with the second. I am not grasping something.

- #9

HallsofIvy

Science Advisor

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The "general" geometric series is a+ ar+ ar

Obviously "a" is just the first term: 2/3. r= ar/r is just the second term divided by the first term: -(4/9)/(2/3)= what?

Now put those into a/(1-r)

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