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Consecutive Integer Series (Posted on 2007-05-14) Difficulty: 3 of 5
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 formulaMindrod2007-05-17 00:33:26
SolutionFull expicit formulaJer2007-05-15 12:07:13
re(3): A formulaDennis2007-05-15 07:57:21
re: NAME ITCharlie2007-05-15 01:30:18
re(2): A formulaCharlie2007-05-15 01:23:01
re(2): A formulaDej Mar2007-05-14 22:28:57
re: A formulaMindrod2007-05-14 21:04:52
NAME ITAdy TZIDON2007-05-14 18:13:48
re: A formulaDennis2007-05-14 16:48:08
re: Better but worseDennis2007-05-14 16:28:10
Some ThoughtsBetter but worseJer2007-05-14 15:36:15
SolutionA formulaDej Mar2007-05-14 15:09:16
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