Summation of series using method of difference


by hhm28
Tags: difference, method, series, summation
hhm28
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#1
Aug17-12, 05:40 AM
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1. The problem statement, all variables and given/known data
Here's my question. My school recently taught me finding summation using method of difference and what my teacher taught was just involving 2 partial fractions.

But this question appeared in my exercise given by my teacher.

r th term: (2r-1)/r(r+1)(r+2). Find summation of n th terms begin with r=1.

Can someone show me how to solve this using method of difference?


2. The attempt at a solution
What I got is sum (- 1/2r + 3/r+1 - 5/2(r+2) ).

And I stucked. What I used to know is r th term is f(r) - f(r-1). and sum of rth term is f(n)- f(0).
I cant get the f(r) - f(r-1) either because -1/2r and -5/2(r+2) both having the same sign.
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tiny-tim
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Aug17-12, 06:03 AM
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hi hhm28! welcome to pf!
Quote Quote by hhm28 View Post
r th term: (2r-1)/r(r+1)(r+2). Find summation of n th terms begin with r=1.
might be easier to split it up first
for example 2/(r+1)(r+2) - 1/r(r+1)(r+2)
hhm28
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#3
Aug17-12, 06:56 AM
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This did help me. But the problem is i stucked somewhere.

I got sigma -[(2/r+2) - (3/r+1) + (1/2r+2) + (1/2r)] and I stucked.

I should've make it into f(r)-f(r-1). =( I'm stupid Argh....

Can you show me how? Perhaps upload a photo. LOL

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Aug17-12, 07:04 AM
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Summation of series using method of difference


hint: how would you do ∑ 1/r(r+1) ?

use the same method for ∑ 1/r(r+1)(r+2)
hhm28
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#5
Aug17-12, 07:09 AM
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LOL. Thats the problem. I only know how to solve f(r) - f(r-1).

Now there is 3 partial fractions. But with 2/(r+1)(r+2) - 1/r(r+1)(r+2), i managed to get ∑ -[(2/r+2) - (3/r+1) + (1/2r+2) + (1/2r)].
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Aug17-12, 07:45 AM
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ok, so how would you do ∑ 1/r(r+1) ?
hhm28
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#7
Aug17-12, 07:56 AM
P: 13
1/r(r+1) = (1/r) - (1/r+1)

∑ 1/r(r+1) = ∑ (1/r) - (1/r+1)
= -∑ [(1/r+1) -(1/r)]

Let f(r)= 1/ r+1 , f(r-1)= 1/r
∑ 1/r(r+1) = -∑ f(r) - f(r-1)
= - [f(n) - f(0)]
= - [(1/n+1) - 1]
= n/n+1
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Aug17-12, 08:10 AM
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Quote Quote by hhm28 View Post
1/r(r+1) = (1/r) - (1/r+1)
ok, now do the same for ∑ 1/r(r+1)(r+2)
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#9
Aug17-12, 08:22 AM
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∑1/r(r+1)(r+2) = ∑1/2r - 1/(r+1) + 1/2(r+2)

How do i continue? the 1/(r+1) is disrupting. =(
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Aug17-12, 08:35 AM
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Quote Quote by hhm28 View Post
∑1/r(r+1)(r+2) = ∑1/2r - 1/(r+1) + 1/2(r+2)
no, try double fractions like 1/r(r+1)
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#11
Aug17-12, 08:43 AM
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means that A/r(r+1) + B/(r+2) to get partial?
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Aug17-12, 08:48 AM
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A/r(r+1) + B/(r+1)(r+2)
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#13
Aug17-12, 08:58 AM
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Gosh. I never learned that =(
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#14
Aug17-12, 09:01 AM
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Thank you so much anywhere. =)
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Aug19-12, 06:37 PM
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Quote Quote by hhm28 View Post
1. The problem statement, all variables and given/known data
Here's my question. My school recently taught me finding summation using method of difference and what my teacher taught was just involving 2 partial fractions.

But this question appeared in my exercise given by my teacher.

r th term: (2r-1)/r(r+1)(r+2). Find summation of n th terms begin with r=1.

Can someone show me how to solve this using method of difference?


2. The attempt at a solution
What I got is sum (- 1/2r + 3/r+1 - 5/2(r+2) ).

And I stucked. What I used to know is r th term is f(r) - f(r-1). and sum of rth term is f(n)- f(0).
I cant get the f(r) - f(r-1) either because -1/2r and -5/2(r+2) both having the same sign.
Good start. Try writing out the first few terms:

r=1: ##-\frac{1}{2}\left(\frac{1}{1}\right) + 3\left(\frac{1}{2}\right) - \frac{5}{2}\left(\frac{1}{3}\right)##

r=2: ##-\frac{1}{2}\left(\frac{1}{2}\right) + 3\left(\frac{1}{3}\right) - \frac{5}{2}\left(\frac{1}{4}\right)##

r=3: ##-\frac{1}{2}\left(\frac{1}{3}\right) + 3\left(\frac{1}{4}\right) - \frac{5}{2}\left(\frac{1}{5}\right)##

r=4: ##-\frac{1}{2}\left(\frac{1}{4}\right) + 3\left(\frac{1}{5}\right) - \frac{5}{2}\left(\frac{1}{6}\right)##

r=5: ##-\frac{1}{2}\left(\frac{1}{5}\right) + 3\left(\frac{1}{6}\right) - \frac{5}{2}\left(\frac{1}{7}\right)##

Now look at the pieces with (1/3) (or (1/4) or (1/5)). Can you see a pattern to how the various parts cancel?


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