# Series, Sequence and Probablility Question

I am working with problems which are taking a toll on me.
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

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HallsofIvy
Homework Helper
Do you understand what a "partial sum" is? That first problem is just asking you to find A1= (-2)1+ 5 (the "first partial sum") and then
A1+ A2+ A3= ((-2)1+5)+ ((-2)2+ 5)+ ((-2)3+ 5).

Yes, the sum of an infinite sum a+ ar+ ar2+ ... is a/(1-r).
Here, you have (2/3)+ (2/3)(-2/3)+ (2/3)(-2/3)2+... What are a and r?

Thanks for the help, I have come up with S1=3 and S3=-3

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

Am I correct?

VietDao29
Homework Helper
Your S1 is correct, but I think you should re-check your answer for S3, S3 = A1 + A2 + A3.
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,

VietDao29 said:
Your S1 is correct, but I think you should re-check your answer for S3, S3 = A1 + A2 + A3.
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,
For the S3= -3
((-2)3+ 5) = -3 (Am I not doing this correctly?)

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

VietDao29
Homework Helper
Nope, you are not doing it correctly.
Sn is the sum of the first n terms.
So S3 is the sum of the first 3 terms. So:
S3 = A1 + A2 + A3 = ...
Note that they are not asking for A3, they are asking for S3.
So what do you get for S3?
--------------------
How can you come up with r = 1 / 3???
$$a_1 = \frac{2}{3}$$
$$a_2 = -\frac{4}{9} = a_1r$$
So again, what is r?
Viet Dao,

Last edited:
Ah ha! I have come up with 9 for S3.

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

HallsofIvy