Series, Sequence and Probablility Question

[TonyC]

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

 

[HallsofIvy]

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?

 

[TonyC]

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]

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,

 

[TonyC]

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

 

[TonyC]

Can anyone lend some advice?

 

[VietDao29]

Nope, you are not doing it correctly.
S_n 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,

 

[TonyC]

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

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

 

[HallsofIvy]

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)