Some Algebra ProblemsDate: 6/1/96 at 21:56:38 From: Anonymous Subject: Series Conjecture but do not prove the formula for 1/2.3 + 1/3.4 + 1/4.5 +...+ n. Show all work. Date: 6/2/96 at 7:37:40 From: Doctor Anthony Subject: Re: Series The sum of this series is easily found using the method of differences. Note first that 1/2.3 = 1/2 - 1/3 1/3.4 = 1/3 - 1/4 1/4.5 = 1/4 - 1/5 ................. ................. 1/n(n+1) = 1/n - 1/(n+1) 1/(n+1)(n+2) = 1/(n+1) - 1/(n+2) ---------------------------------- Add all the equations. The lefthand side gives the series we want, and on the righthand side we note that terms cancel between lines, i.e. -1/3 in the first line cancels with +1/3 in the second line. In fact every term cancels except the first term on the first line and the last term on the last line. Sum of series = 1/2 - 1/(n+2) We note from this that the sum to infinity would be 1/2. If we add the two terms we get sum of series = n/{2(n+2)} -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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