# Homework Help: Find the sum of a series.

1. Feb 15, 2012

### naaa00

1. The problem statement, all variables and given/known data

Hello there!

I'm trying to find the sum of:

1 - 3 + 3^2 - 3^3 + ... + 3^(2n) = ...

3. The attempt at a solution

Well, I tried with

1 - x + x^2 - x^3 + ... = (1 - x^n)/(1 - x)

with x = 3. However, the nth term doesn't fit (3^(2n)).

At the moment, I'm a little bit out of ideas. Any suggestion would be very appreciated.

2. Feb 15, 2012

### Dick

This formula is wrong. Why don't you look up the correct one to get started?

3. Feb 15, 2012

### naaa00

Oh yeah, my bad.

I meant

1 + x + x^2 + x^3 + ... = (1 - x^n)/(1 - x)

4. Feb 15, 2012

### Dick

Still not good. There should be a last term on the left. It's not an infinite series.

5. Feb 15, 2012

### naaa00

1 + x + x^2 + x^3 + ... + x^(n-1) = (1 - x^n)/(1 - x)

I'm currently trying, but still nothing.

6. Feb 15, 2012

### Dick

That's it. Let's write that as 1 + x + x^2 + x^3 + ... + x^(m-1) = (1 - x^m)/(1 - x). It looks to me like you just want x=(-3) and m-1=2n.

7. Feb 15, 2012

### naaa00

I'm trying for example:

The nth term 3^(2n), I'm wrinting it as (3^n)^2

Then I got: (Sigma 3^n) times (Sigma 3^n). Each one is a geometric series and the sum for each would be (1 - 3^n)/(1 - 3). Since there is a product of geometric series, I supposed that on the RHS I could put (1 - 3^n)/(1 - 3) times (1 - 3^n)/(1 - 3).

Or [(1 - 3^n)/(1 - 3)]^2

But apparently it doesn't work...

8. Feb 15, 2012

### Dick

No. It's a single geometric series. You hardly have to do anything except substitute into a formula. 3^(2n)=(-3)^(2n). Reread my last post.

9. Feb 15, 2012

### naaa00

Omg. I think my mistake was that I used all the time x = 3 and not x =-3!

Let's see.

10. Feb 15, 2012

### naaa00

...

Let's see:

m- 1 = 2n, m = 2n + 1, and x = -3,

(1 - (-3^{2n+1})/ 4

For n = 0, 1.

[1 - (-3^1)]/4 = 1, [1 - (-3^3)]/4 = 28/4 = 7.

but the sum of the first two terms is -2, and not 7.

1 - 3 + 3^2 - 3^3 + ... + 3^(2n)

Apparently doesn't work, (or its me?) Why it doesn't work?

11. Feb 15, 2012

### Dick

If you put n=1 then you are summing up to 3^(2*n)=3^2=9. That's the sum of the first three terms 1-3+9. Not the first two terms.

12. Feb 15, 2012

### naaa00

Aha! Now I get it. Thank you very much for your help!

13. Feb 15, 2012

### naaa00

Now, I'm trying to prove this by induction, but I doesn't work...

[1 - x^{2n+1}/4] + 3^{2n+1} = [1 - x^{2n+1} + 4x^{2n+1}]/4

[1 - x^{2n+1} (1 + 4)]/4 = [1 - 5x^{2n+1}]/4,

susbtituting x =-3, [1 - 5(-3)^{2n+1}]/4,

but, I should get [1 - (-3)^{2n+3}/4]

Any suggestion?

14. Feb 15, 2012

### Dick

The n case is the sum up to (-3)^(2n). To get to the n+1 case, which sums up to (-3)^(2(n+1))=(-3)^(2n+2) you need to add two terms. (-3)^(2n+1) and (-3)^(2n+2).

Last edited: Feb 15, 2012
15. Feb 15, 2012

### naaa00

Yeah, indeed, it works.

I must admit I don't fully understand why for the n+1 term, I must add two terms. Why is that?

16. Feb 15, 2012

### naaa00

So, from the n+1 term: (-3)^(2(n+1))=(-3)^(2n+2)

the distance from n, is 2n+2, and therefore I must include the number which is in between, that is, 2n+1?

17. Feb 15, 2012

### Dick

Yes. The case n and the case n+1 differ by the addition of two terms.

18. Feb 15, 2012

### naaa00

Ok. Thanks again!

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