All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Consecutive Integer Series (Posted on 2007-05-14)
In the infinite series 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 4, 5... each positive integer k appears k times in consecutive order.

Write a formula for the sum of the first n terms of the series.

 Submitted by Dennis Rating: 4.5000 (2 votes) Solution: (Hide) Let s(k) denote the sum of the first k terms of the series. Now s(1)=1^2, s(3)=1^2+2^2, and in general s(k(k+1)/2)=1^2 + 2^2 + . . . + k^2 --> s(k(k+1)/2) = k(k+1)(2k+1)/6 Letting n=k(k+1)/2 --> s(n) = n*sqrt(8n+1)/3 when 8n+1 is a perfect square and n is a triangular number. Let m denote the largest triangular number less than or equal to n. So s(m)=m*sqrt(8m+1)/3 and m=j(j+1)/2 for some positive integer j. Now m is the j_th triangular number --> j^2 + j - 2m = 0 --> j = (-1 + sqrt(8m+1))/2 --> j = [ (-1+sqrt(8n+1))/2 ] where [w] denotes the greatest integer less than or equal to w. But s(n) = s(m) + (n-m)(j+1). Substituting the above equations for s(m) and m, and then simplifying yields s(n) = (j + 1)(6n - j^2 - 2j)/6.

Comments: ( You must be logged in to post comments.)
 Subject Author Date re(3): A formula Mindrod 2007-05-17 00:33:26 Full expicit formula Jer 2007-05-15 12:07:13 re(3): A formula Dennis 2007-05-15 07:57:21 re: NAME IT Charlie 2007-05-15 01:30:18 re(2): A formula Charlie 2007-05-15 01:23:01 re(2): A formula Dej Mar 2007-05-14 22:28:57 re: A formula Mindrod 2007-05-14 21:04:52 NAME IT Ady TZIDON 2007-05-14 18:13:48 re: A formula Dennis 2007-05-14 16:48:08 re: Better but worse Dennis 2007-05-14 16:28:10 Better but worse Jer 2007-05-14 15:36:15 A formula Dej Mar 2007-05-14 15:09:16

 Search: Search body:
Forums (0)