Number series help needed

J

jimi

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Feb 1, 2005
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(a) "Sum the series 1 + x + x^2 + ... + x^n for x does not equal 1."

I know how to work this out but I just wanted to check this: is this the sum of n+1 terms?

(b) "Find the sum of the series 4/3 + 9/8 + 16/15 + ... + n^2/(n^2-1)"

I'm okay working this out too but again just wanted to check this: is this the sum of n-1 terms?

I'm clear when the question asks for the sum of n terms but find it a little confusing when it's something other than n.
Thanks for any help.
 
(a) Just look at the exponents. 0, 1, 2,...n makes n+1 terms. There are a couple of interesting ways to add up finite geometric series. How are you doing it?
 
g.p. sum of n+1 terms = a(1-r^(n+1))/(1-r) where a is the first term and r is the common ratio. Here, a = 1 and r = x so:
Sn+1= (1-x^(n+1))/(1-x)
 
Ah, you have the magic formula. That will work. Good job.
 
You're correct about 2 thru n being n-1 terms. Now embarrass me. Did you get it to be a geometric series? It surely doesn't look arithmetric! I'm missing the magic :oops:
----------------
Gene
 
Gene
It's not a g.p. or an a.p.
I used the 'method of differences':

Firstly, I changed n^2/(n^2-1) to 1 + 1/(2(n-1)) - 1/(2(n+1))

1st term (n=2): 1 + 1/2 - 1/6
2nd term (n=3): 1 + 1/4 - 1/8
3rd term (n=4): 1 + 1/6 - 1/10
4th term (n=5): 1 + 1/8 - 1/12

Sum = (n-1) + 1/2 + 1/4 - 1/(2n) - 1/(2(n+1))
= ((n-1)(4n^2 + 7n + 2))/(4n(n+1))
 
Gee, that's something I didn't have in my arsenal. Thanks.
------------------
Gene
 
jimi said:
1st term (n=2): 1 + 1/2 - 1/6
2nd term (n=3): 1 + 1/4 - 1/8
3rd term (n=4): 1 + 1/6 - 1/10
4th term (n=5): 1 + 1/8 - 1/12
Click to expand...
Well, keep it simple: change n to equal the term number, drop the 1:
1st term (n=1): 1/2 - 1/6
2nd term (n=2): 1/4 - 1/8
3rd term (n=3): 1/6 - 1/10
4th term (n=4): 1/8 - 1/12

Now you have 1/2n - 1/(2n+4)
Simplifies to 1 / [n(n+2)]
Example, 4th term: 1/ [4(6)] = 1/24 = 1/8 - 1/12

To include the 1: 1 + 1/[n(n+2)]

First few:
1 + 1/3
1 + 1/8
1 + 1/15
1 + 1/24
1 + 1/35
1 + 1/48
and so on...
Notice the progression in the denominators: 5,7,9,11,13.....

Now don't tell me that's not geometric !
 
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