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I got this exercise from school and I have no idea what notion to use, it resumes to Harmonic series, I can't find a generic answer. Do you have any idea?

$\frac{6}{4 \times 2} + \frac{7}{5 \times 2} + ... + \frac{21}{19 \times 2}$

Which is the generic answer for such a sum?

minseong
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yonutix
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  • What is the "*"? Is it the usual multiplication operation or something else? Please explain if you want an answer :) – Yes May 03 '15 at 10:14
  • multiplication yes – yonutix May 03 '15 at 10:15
  • What is the question? – Bernard May 03 '15 at 10:15
  • Which is the generic answer, how can this sum be computed when I have 100 elements or more. There is a solution? – yonutix May 03 '15 at 10:16
  • Note that for each $k$, the $k$th term is simply $$\frac{1}{2} + \frac{1}{k+3}.$$ – Yes May 03 '15 at 10:18
  • $ \LARGE{\frac{6}{42} + \frac{7}{52} + ... + \frac{21}{19*2} = \ = \frac{1}{2} (\frac{6}{4} + \frac{7}{5} + ... + \frac{21}{19} ) = \ = \frac{1}{2}(16 + [\frac{2}{4} + \frac{2}{5} + ... + \frac{2}{19}]) = \ = 8 + (\frac{1}{4} + \frac{1}{5} + \cdots \frac{1}{19}) }$ – Tahir Imanov May 03 '15 at 10:21
  • Yes, this is a harmonic serie, but how can I solve the parathesis? – yonutix May 03 '15 at 10:26

2 Answers2

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Consider the most general case $$A=\sum_{i=a}^b\frac n{2(n-2)}=\frac 12\sum_{i=a}^b\frac n{n-2}=\frac 12\sum_{i=a}^b\frac {n-2+2}{n-2}$$ $$A=\frac 12\sum_{i=a}^b1+\sum_{i=a}^b\frac {1}{n-2}=(b-a+1)+\sum_{i=a}^b\frac {1}{n-2}$$ So, you are just left with the sum of fractions $\frac 14+\frac 15+\frac 16+\cdots+\frac 1{19}$

Claude Leibovici
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We got the following sum:

$$\sum\limits _{n=6}^{21}\frac{n}{2(n-2)}=$$

$$\frac{6}{2(6-2)}+\frac{7}{2(7-2)}+\frac{8}{2(8-2)}+\frac{9}{2(9-2)}+...+\frac{21}{2(21-2)}=$$

$$\frac{6}{8}+\frac{7}{10}+\frac{8}{12}+\frac{9}{14}+...+\frac{21}{38}=$$

$$\frac{753813839}{77597520}$$

Jan Eerland
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