Based on: $\frac{1}{n*(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ where n is element of N find the sum of the following:

$\frac{1}{1*2}+\frac{1}{2*3}+\frac{1}{3*4}+ ... +\frac{1}{38*39}+\frac{1}{39*40}$

How should one deal with this kind of problem? Is this a mathematical induction, arithmetic series, geometric series? I'm lost on this one.

Here are the options: a)$\frac{31}{40}$ b)$\frac{33}{40}$ c)$\frac{37}{40}$ d)$\frac{39}{40}$


marked as duplicate by Did sequences-and-series May 14 '17 at 11:56

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    $\begingroup$ It's telescoping. The required decomposition has already been provided; what you then find is that terms cancel out everywhere. $\endgroup$ – Parcly Taxel May 14 '17 at 11:39
  • $\begingroup$ Try to expand a bit and observe the telescoping : $$ \frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} = (\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} -\frac{1}{4}) $$ $\endgroup$ – Zubzub May 14 '17 at 11:44

$\frac{1}{1*2}+\frac{1}{2*3}+ ... +\frac{1}{38*39}+\frac{1}{39*40}=\\ (\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+...+(\frac{1}{38}-\frac{1}{39})+(\frac{1}{39}-\frac{1}{40})=\\ \frac{1}{1}+(-\frac{1}{2}+\frac{1}{2})-\frac{1}{3}+...+\frac{1}{38}+(-\frac{1}{39}+\frac{1}{39})-\frac{1}{40}=\\ \frac{1}{1}-\frac{1}{40}$

Can you see that all terms except for the first and last term cancel?


Since each and every term will be canceled out except $1$ and $-\dfrac{1}{40}$

So, the answer is $$1-\frac{1}{40}=\frac{39}{40}$$

See this answer.

Hope it helped!!

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    $\begingroup$ The thing is I never learned about telescoping series, when I saw this problem i was confused, it is the first time i'm hearing about telescoping and maybe when i learn about these series it might be easy to solve this problem. $\endgroup$ – L.B May 14 '17 at 11:48
  • $\begingroup$ See the link I have provided it will help you to understand this series's sum. $\endgroup$ – Harsh Kumar May 14 '17 at 11:50
  • $\begingroup$ @L.B , Also see en.wikipedia.org/wiki/Telescoping_series $\endgroup$ – Harsh Kumar May 14 '17 at 12:08
  • $\begingroup$ It is very helpful thank you! $\endgroup$ – L.B May 14 '17 at 12:11

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