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Summing Consecutive Terms and Pair Determination (Posted on 20151121) 

The sequence S_{1}, S_{2}, S_{3}, ..... is defined by:
S_{k} = 1/(k^{2} + k), for k=1,2,.....
Find M and N, given that:
S_{M}+ S_{M+1}+ .....+S_{N} = 1/29
Solution

 Comment 1 of 3

1/(k^{2}
+ 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/(k^{2}
+ 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 20151121 11:20:56 


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