
#1
Aug1712, 05:40 AM

P: 13

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: (2r1)/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(r1). and sum of rth term is f(n) f(0). I cant get the f(r)  f(r1) either because 1/2r and 5/2(r+2) both having the same sign. 



#2
Aug1712, 06:03 AM

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hi hhm28! welcome to pf!
for example 2/(r+1)(r+2)  1/r(r+1)(r+2) 



#3
Aug1712, 06:56 AM

P: 13

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(r1). =( I'm stupid Argh.... Can you show me how? Perhaps upload a photo. LOL 



#4
Aug1712, 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) 



#5
Aug1712, 07:09 AM

P: 13

LOL. Thats the problem. I only know how to solve f(r)  f(r1).
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)]. 



#7
Aug1712, 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(r1)= 1/r ∑ 1/r(r+1) = ∑ f(r)  f(r1) =  [f(n)  f(0)] =  [(1/n+1)  1] = n/n+1 



#8
Aug1712, 08:10 AM

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#9
Aug1712, 08:22 AM

P: 13

∑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. =( 



#10
Aug1712, 08:35 AM

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#11
Aug1712, 08:43 AM

P: 13

means that A/r(r+1) + B/(r+2) to get partial?




#13
Aug1712, 08:58 AM

P: 13

Gosh. I never learned that =(




#14
Aug1712, 09:01 AM

P: 13

Thank you so much anywhere. =)




#15
Aug1912, 06:37 PM

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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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