Nth Term of a SeriesDate: 08/27/97 at 07:45:01 From: Alice Subject: Question! 1/(1*2*3)+1/(2*3*4)+.......+1/<k(k+1)(k+2)>+.......+1/(100*101*102) =? Thanks! Date: 08/27/97 at 08:46:40 From: Doctor Anthony Subject: Re: Question! 1 The nth term is of the form -------- n(n+1)(n+2) 1 1 We can split this into (1/2)[ ----- - ----------] n(n+1) (n+1)(n+2) The terms of the series can be written: nth term = (1/2)[1/n(n+1) - 1/(n+1)(n+2)] (n-1)th term = (1/2)[1/(n-1)n - 1/n(n+1) ] (n-2)th term = (1/2)[1/(n-2)(n-1) - 1/(n-1)n ] ------------ ---------------------------------- ------------ ---------------------------------- 3rd term = (1/2)[1/((3)(4)) - 1/((4)(5))] 2nd term = (1/2)[1/((2)(3)) - 1/((3)(4))] 1st term = (1/2)[1/((1)(2)) - 1/((2)(3))] ----------------------------------------------------- Add all terms Total = (1/2)[1/((1)(2)] - (1/2)[1/(n+1)(n+2)] These are the only terms that remain. All the others cancel between rows. The formula is SUM = (1/2)[1/2 - 1/(n+1)(n+2)] If n = 100 this gives SUM = (1/2)[1/2 - 1/(101)(102)] = 1/4 - 1/20604 = 2575/10302 (=.249951465..) The series converges to the value 1/4 as n -> infinity. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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