# Sequences and series

1. Sep 6, 2010

### ings

CAN U ALL HELP ME TO SOLVE THIS QUESTION?
I don't know how to start.....

The sum of the first 2n terms of a series P is 20n-4n2. Find in terms of n, the sum of the first n terms of this series. Show that the series is an arithmetic series.

2. Sep 7, 2010

### Mentallic

Well if the sum of 2n terms is .... then what they gave you is equal to that. Now the formula for the sum of n terms is ... and you have to substitute that into the sum of 2n terms formula by manipulating a few things.

3. Sep 7, 2010

### ings

But the question only mentions that sentences ...

4. Sep 7, 2010

### Mentallic

Hmm... now that I read it again I was planning on assuming the series is arithmetic. But even then I can't seem to isolate Sn without having d in the equation. Which class are you in?

5. Sep 7, 2010

### betel

If I assume it is an arithmetic series the series is uniquely determined.

But I don't think the statement above is enough. I could easily fix an arbitrary series of number that give the same sum for the first 2n terms. We have two numbers to choose ($$a_{2n+1}, a_{2n+2}$$) and only one constraint ($$a_{2n+1}+a_{2n+2}=16-8n=S_{2n+2}-S_{2n}$$).

Last edited: Sep 7, 2010
6. Sep 7, 2010

### HallsofIvy

Staff Emeritus
So the "0" th term is 0, the sum of the first 2 terms is 20(1)- 4(1)= 16, the sum of the first 4 terms is 20(2)- 4(4)= 24, the sum of the first 6 terms is 20(3)- 4(9)= 24, the sum of the first 8 terms is 20(4)- 4(16)= 16, the sum of the first 10 terms is 20(5)- 4(25)= 0, etc. In order to go up from 0 to 16 in two steps, the series must have a positive difference. But after the 6th term, it is decreasing so must have a negative difference. I don't see how that can be an arithmetic sequence that must have a constant difference.

7. Sep 7, 2010

### uart

I'm pretty sure that this is a basic "school type" problem that *assumes* that all series are either Arithmetic or Geometric. A geometric series will have a sum that is an exponential function of "n" and an arithmetic series will have a sum that's a quadratic function of "n". So out of those two options it's clearly Arithmetic.

In terms of the first term "a" and the common difference "d" the sum of an AS is :

S(n) = (a-d/2) n + d/2 n^2

so S(2n) = (2a-d) n + 2d n^2, and the rest is very straight foward.

8. Sep 7, 2010

### vela

Staff Emeritus
You don't need to assume anything about the type of sequence. The key is to find the expression for Sn first, which you can deduce from S2n, and then calculate an=Sn-Sn-1. You'll find that an is an arithmetic sequence.

9. Sep 7, 2010

### betel

I don't think you can deduce $$S_n$$
I can do arbitrary series to get the result, e.g.
$$a_n=(1+(-1)^n)(12-2n)$$
will give the requested sum, but is not an arithmetic series, as every odd term is 0.
The first terms in this series are
$$a_1=0,a_2=16,a_3=0,a_4=8,a_5=0,a_6=0,a_7=0,a_8=-8,a_9=0$$
And the sum of the first 2n terms is given by the formula above.

10. Sep 7, 2010

### vela

Staff Emeritus
I guess it depends on whether you interpret "the sum of the first 2n terms" to mean S2n, in which case, it's straightforward to find Sn, or you interpret it to mean that f(n)=20n-4n2 happens to equal S2n for n=0,1,....

11. Sep 7, 2010

### betel

I don't get the difference on that

12. Sep 7, 2010

### vela

Staff Emeritus
It's the difference between figuring out f(x) if you know what f(2x) as opposed to figuring out f(x) if all you know is f(x)=g(x) for x=0, 1, 2, ....

Last edited: Sep 7, 2010
13. Sep 7, 2010

### ings

i am not sure whether it's correct...

i do it in this way: let S2n= f(2n)
then Sn=f(n)

f(2n)= 20n-4n2
= 10(2n)- (2n)2

hence, f(n)= 10n - n2
SO Sn = 10n - n2

Is it correct?

Last edited: Sep 7, 2010
14. Sep 9, 2010

### Redbelly98

Staff Emeritus
Yes, that looks good. Next, try using vela's advice in Post #8 to find an.

Last edited: Sep 9, 2010
15. Sep 9, 2010

### Redbelly98

Staff Emeritus
Moderator's note:

FYI, I have deleted some posts that I felt were confusing to, and ignoring, the student. Members, please consider using the forum's Private Message system to discuss and clarify details about the question with each other.

As this thread has been moved to the Homework & Coursework Questions forums since its original posting, the usual rules about homework help now apply.

Last edited: Sep 9, 2010
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