Decomposing FractionsDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/