Difference method with partial fractions

In summary, the conversation discusses verifying a mathematical equation and finding the sum of a series using the difference method. The key issue in using the difference method is properly classifying the terms in the equation.
  • #1
look416
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Homework Statement



Verify that, for all positive values of n,
1/(n+2)(2n+3) -1/((n+3)(2n+5))=(4n+9)/((n+2)(n+3)(2n+3)(2n+5))
For the series
∑_(n=0)^N▒(4n+9)/((n+2)(n+3)(2n+3)(2n+5))
Find
The sum to N terms,
The sum to infinity.


Homework Equations



no

The Attempt at a Solution



i had solved the first sextion which is to verification
however, when it came to the difference method solving, i had a huge problems
therefore, any solutions?
 
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  • #3
most of the google doesn't reveal the method of solving series using difference method ==
i have tried it
or maybe you can give me the links if you have found it
 
  • #4
[tex]\frac{1}{(n+2)(2n+3)} = \frac{2}{2n+3} - \frac{1}{n+2}[/tex]
[tex]\frac{1}{(n+3)(2n+5)} = \frac{2}{2n+5} - \frac{1}{n+3}[/tex]
I think this may be the key issue bugging you when trying to use the difference method without simplifying the expressions further.
 
  • #5
Fightfish said:
[tex]\frac{1}{(n+2)(2n+3)} = \frac{2}{2n+3} - \frac{1}{n+2}[/tex]
[tex]\frac{1}{(n+3)(2n+5)} = \frac{2}{2n+5} - \frac{1}{n+3}[/tex]
I think this may be the key issue bugging you when trying to use the difference method without simplifying the expressions further.

well i have found the answer
thanks for the help
my fault is i did not classified them correctly
as i should put 1/(n+3) and 1/(n+2) together as well as 1/(2n+3) and 2/(2n+5) together instead of i straight use the result to do difference method that's y i can't find the answer!
Anyways thanks for your help
 

FAQ: Difference method with partial fractions

What is the difference method with partial fractions?

The difference method with partial fractions is a mathematical technique used to simplify and solve integrals of rational functions. It involves breaking down the rational function into simpler fractions, and then using the difference method to integrate each fraction separately.

When is the difference method with partial fractions used?

This method is typically used when integrating a rational function that cannot be easily solved using other techniques, such as substitution or integration by parts. It is also commonly used in solving differential equations.

How does the difference method with partial fractions work?

The first step is to factor the denominator of the rational function into linear and irreducible quadratic factors. Then, using the method of undetermined coefficients, the coefficients of the partial fractions are determined. Finally, the difference method is used to integrate each fraction separately.

What are the benefits of using the difference method with partial fractions?

This method allows for the integration of complex rational functions by breaking them down into simpler fractions. It also provides a systematic approach to solving integrals of rational functions, making it easier to follow and less prone to errors.

Are there any limitations to the difference method with partial fractions?

One limitation is that it can only be used for rational functions. It also may not work for all rational functions, as some may require alternative techniques for integration. Additionally, it can be time-consuming and tedious for more complex rational functions.

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