Series, Sequence and Probablility Question

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SUMMARY

The discussion focuses on calculating partial sums of the sequence defined by An=(-2)^n+5 and the sum of a geometric series starting with 2/3. Participants clarify that the first partial sum, S1, is correctly calculated as 3, while S3 requires summing the first three terms, leading to a correct value of 9. The geometric series formula S=A1/(1-r) is confirmed, with participants identifying the first term (a) as 2/3 and the common ratio (r) as -2/3, leading to further calculations for the infinite series.

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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
Please help :eek:
 
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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?
 
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
 
Can anyone lend some advice?
 
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? :smile:
--------------------
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.
 
Your sum is 2/3 - 4/9 + 8/27 -...

The "general" geometric series is a+ ar+ ar2+ ar3+...

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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