perplexus.info :: Sequences : Summing Consecutive Terms and Pair Determination

Summing Consecutive Terms and Pair Determination (Posted on 2015-11-21)

The sequence S₁, S₂, S₃, ..... is defined by:

S_k = 1/(k² + k), for k=1,2,.....

Find M and N, given that:

S_M+ S_M+1+ .....+S_N = 1/29

See The Solution Submitted by K Sengupta

Rating: 5.0000 (1 votes)

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Solution

| Comment 1 of 3

1/(k² + k) = 1/k – 1/(k+1), so consecutive terms have
parts that cancel as follows:

1/2 + 1/6 + 1/12 + 1/20 + …
= (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1/4) + (1/4 – 1/5) +…

Thus: Sum_{k=M to N}(1/(k² + k)) = 1/M – 1/(N + 1)

which gives: 1/M – 1/(N + 1) = 1/29

(29 – M)(N + 1) = 29M

So M < 29 and, since 29 is prime, neither 29 nor M can have
prime factors in common with 29 – M.

Thus 29 – M = 1, giving M = 28 and N = 811

Posted by Harry on 2015-11-21 11:20:56

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