 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Summing Consecutive Terms and Pair Determination (Posted on 2015-11-21) The sequence S1, S2, S3, ..... is defined by:

Sk = 1/(k2 + k), for k=1,2,.....

Find M and N, given that:

SM+ SM+1+ .....+SN = 1/29

 No Solution Yet Submitted by K Sengupta No Rating Comments: ( Back to comment list | You must be logged in to post comments.) 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:  Sumk=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 Please log in:
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