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

Home > Numbers > Sequences
Sum 2001 Terms (Posted on 2016-02-27) Difficulty: 3 of 5
Each term of the sequence {S(n)} is an integer, where:
S(n+2) = S(n+1) – S(n) for n > 0

The sum of the first 1492 terms is 1985, and:
The sum of the first 1985 terms is 1492.

Find the sum of the first 2001 terms.

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution Comment 1 of 1
Let S(1) = a and S(2) = b.  Then S(3) = b-a,  S(4) = -a, S(5) = -b, S(6) = a-b, S(7) = a, S(8) = b, etc.  The sequence is cyclic mod 6, also the sum of any 6*n consecutive terms is zero.

Sum{S(1) to S(1492)} = S(1) + S(2) + S(3) + S(4) + Sum{S(5) to S(1492)} = a + b + b-a + -a + 0 = 2b-a = 1985

Sum{S(1) to S(1985)} = S(1) + S(2) + S(3) + S(4) + S(5) + Sum{S(6) to S(1985)} = a + b + b-a + -a + -b + 0 = b-a = 1492

Then b=493 and a = -999

Sum{S(1) to S(2001)} = S(1) + S(2) + S(3) + Sum{S(4) to S(2001)} = a + b + b-a + 0 = 2b = 986

  Posted by Brian Smith on 2016-02-27 11:18:03
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (7)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information