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Decomposing Fractions

Date: 04/08/2009 at 09:34:35
From: Ray
Subject: adding fractions

Find the sum of (1/1x2), (1/2x3), (1/3x4), (1/4x5), 
(1/5x6).............(1/28x29), (1/29x30)

I can't figure out the short method behind this.  I tried grouping
into 1/2  or to add 1st and last fraction, but to no avail.



Date: 04/08/2009 at 10:46:14
From: Doctor Greenie
Subject: Re: adding fractions

Hi, Ray --

Each of the fractions in this sum is of the form

     1
  ------
  n(n+1)

We can find a quick way to find the sum by "decomposing" the given
fraction into "partial fractions" which have simpler denominators.

I was surprised not to find any good examples of partial fraction 
decomposition in the Dr. Math archives, or even on some of my 
favorite other math sites.  So I will use your problem as an example 
to demonstrate the process.

Basically, we want to write the fraction

     1
  ------
  n(n+1)

as the sum of two simpler fractions.  If we add two fractions with 
denominators "n" and "n+1", then the common denominator for the 
resulting fraction will be n(n+1).  So we start the process by 
writing

     1      A      B
  ------ = --- + -----
  n(n+1)    n     n+1

and then we use basic algebraic techniques to find the values of A 
and B by adding the two fractions on the right using the common 
denominator:

   A      B     A(n+1)     Bn     (A+B)n+A
  --- + ----- = ------ + ------ = --------
   n     n+1    n(n+1)   n(n+1)    n(n+1)

We want to have

  (A+B)n + A      1      0n+1
  ---------- = ------ = ------
    n(n+1)     n(n+1)   n(n+1)

This means we must have--by equating coefficients of the "n" terms 
and the constant terms in the two fractions--

  A+B = 0
  A = 1

or

  A = 1;  B = -1

So now we know

     1      1      -1
  ------ = --- + -----
  n(n+1)    n     n+1

or

     1      1      1
  ------ = --- - -----
  n(n+1)    n     n+1

When we apply this pattern to solving your problem, we find

  1/(1*2) + 1/(2*3) + 1/(3*4) + ... + 1/(29*30) =

  (1/1 - 1/2) + (1/2 - 1/3) + ... + (1/28 - 1/29) + (1/29 - 1/30)

We can see that the first fraction in each pair is the same as the 
second fraction in the previous pair, but with the opposite sign.  
So all those fractions will cancel each other out, and our final 
result will contain just the first term in the first pair and the 
second term in the last pair:

  1/(1*2) + 1/(2*3) + 1/(3*4) + ... + 1/(29*30) = 1/1 - 1/30 = 29/30

I hope this helps.  Please write back if you have any further 
questions about any of this.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Polynomials

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