perplexus.info :: Sequences : Sum 2001 Terms

Sum 2001 Terms (Posted on 2016-02-27)

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.

See The Solution Submitted by K Sengupta

Rating: 5.0000 (1 votes)

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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

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