Summation of series using method of difference

In summary: Yes! It looks like the terms that are multiples of 3 add together, while the terms that are multiples of 1 add together and the term that is a multiple of both adds together.In summary, my teacher taught me to find summation using method of difference, but this problem appeared in my exercise. I was not able to solve it using that method, so I tried using the same method for ∑ 1/r(r+1) . Thats when I ran into trouble. I only know how to solve f(r) - f(r-1). I then found the pieces with (1/3) (or (1/4) or (1/5)). Can you see
  • #1
hhm28
13
0

Homework Statement


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 can't get the f(r) - f(r-1) either because -1/2r and -5/2(r+2) both having the same sign.
 
Physics news on Phys.org
  • #2
welcome to pf!

hi hhm28! welcome to pf! :smile:
hhm28 said:
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) :wink:
 
  • #3
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
 
  • #4
hint: how would you do ∑ 1/r(r+1) ?

use the same method for ∑ 1/r(r+1)(r+2) :smile:
 
  • #5
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)].
 
  • #6
ok, so how would you do ∑ 1/r(r+1) ? :smile:
 
  • #7
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
 
  • #8
hhm28 said:
1/r(r+1) = (1/r) - (1/r+1)

ok, now do the same for ∑ 1/r(r+1)(r+2) :smile:
 
  • #9
∑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
hhm28 said:
∑1/r(r+1)(r+2) = ∑1/2r - 1/(r+1) + 1/2(r+2)

no, try double fractions like 1/r(r+1) :wink:
 
  • #11
means that A/r(r+1) + B/(r+2) to get partial?
 
  • #12
A/r(r+1) + B/(r+1)(r+2) :smile:
 
  • #13
Gosh. I never learned that =(
 
  • #14
Thank you so much anywhere. =)
 
  • #15
hhm28 said:

Homework Statement


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 can't 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?
 
Last edited:

What is the method of difference in summation of series?

The method of difference is a technique used to calculate the sum of a series by taking differences between consecutive terms and finding a pattern. This method is especially useful for series with terms that are easily computable, such as arithmetic or geometric series.

How is the method of difference used to find the sum of a series?

To use the method of difference, you first need to find a pattern in the differences between consecutive terms. Then, you can use this pattern to create a general formula for the differences and ultimately, the sum of the series.

What types of series can the method of difference be applied to?

The method of difference can be applied to various types of series, including arithmetic, geometric, and other simple series. However, it may not be as effective for more complex series with irregular terms.

What are the advantages of using the method of difference for summation of series?

The method of difference can be a quick and efficient way to find the sum of a series, especially for simple series. It also allows for easy calculation of partial sums and can be generalized for larger series with similar patterns.

Are there any limitations to using the method of difference for summation of series?

While the method of difference can be useful for many series, it may not work for all types of series. Complex series with irregular terms or patterns may not be easily solved using this method. Additionally, the method of difference may not always provide an exact solution and may require rounding or approximation.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
8
Views
540
  • Precalculus Mathematics Homework Help
Replies
15
Views
525
  • Precalculus Mathematics Homework Help
Replies
6
Views
739
  • Precalculus Mathematics Homework Help
Replies
2
Views
799
  • Calculus and Beyond Homework Help
Replies
2
Views
695
  • Precalculus Mathematics Homework Help
Replies
6
Views
2K
  • Quantum Physics
Replies
1
Views
538
  • Precalculus Mathematics Homework Help
Replies
2
Views
1K
  • Precalculus Mathematics Homework Help
Replies
6
Views
1K
Back
Top