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In my task, I've some questions to solve but I am stuck with the following one.

Question: Find the following sum: $\frac{1}{22–1} + \frac{1}{42 –1} + \frac{1}{62 –1} + \dots + \frac{1}{202–1}.$

I tried to find many patterns but couldn't get one to solve it. Like I know the difference is $20$ in each, but there is not any way to find square and make combination, so I am totally stuck with it.

Can anyone guide me with this??

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    $\begingroup$ Where does this question come from, please? What's the context? $\endgroup$ Commented Sep 7, 2020 at 13:11
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    $\begingroup$ Wolfram finds $$\sum_{n=1}^{10} \frac{1}{20n+1}=\frac{32423431590702190}{227186523709446609}$$ $\endgroup$
    – K.defaoite
    Commented Sep 7, 2020 at 13:11
  • $\begingroup$ @K.defaoite provided answer does not match with yours.. $\endgroup$ Commented Sep 7, 2020 at 16:28
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    $\begingroup$ I don't know what you want, Null. You say the answer to your question is $10/21$, but several of us have told you that it can't possibly be $10/21$. You refuse to say anything about where the problem came from, or where these alleged links are that say it's $10/21$. What are you expecting to get out of this? The one thing you won't see is a proof that the numbers, the way you have written them, add up to $10/21$, because they don't. So, what do you want? $\endgroup$ Commented Sep 9, 2020 at 1:46
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    $\begingroup$ OK. Now: please, please, please where did the problem come from? $\endgroup$ Commented Sep 10, 2020 at 11:23

2 Answers 2

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To get an approximation, write $$S_n=\sum_{k=0}^n \frac{1}{20 k+21}=\frac 1 {20}\sum_{k=0}^n \frac{1}{ k+\frac{21}{20}}\sim \frac 1 {20}\sum_{k=0}^n \frac{1}{ k+1}=\frac 1 {20}H_{n+1}$$ For $n=9$, this would give $$S_{9}=\frac 1 {20}\times\frac{7381}{2520}=\frac{7381}{50400}\approx 0.146448$$ instead of $$S_{9}=\frac{32423431590702190}{227186523709446609}\approx 0.142717$$

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    $\begingroup$ @Tortar. Thanks for the edit ! Cheers :-) $\endgroup$ Commented Sep 7, 2020 at 14:18
  • $\begingroup$ but given correct answer is 10/21. (0.476) but it gives 0.1427. $\endgroup$ Commented Sep 7, 2020 at 16:29
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    $\begingroup$ @NullPointer. It is not ! The correct answer is the monster given twice to you. $\endgroup$ Commented Sep 7, 2020 at 18:36
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Note that $${2\over n^2-1}={1\over n-1}-{1\over n+1}$$ So $${2\over2^2-1}+{2\over4^2-1}+\cdots+{2\over20^2-1}=1-{1\over3}+{1\over3}-{1\over5}+\cdots+{1\over19}-{1\over21}=1-{1\over21}={20\over21}$$ and we're done (if I've correctly guessed what question OP meant to ask).

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  • $\begingroup$ good interpretation skills ahah $\endgroup$
    – Tortar
    Commented Sep 8, 2020 at 11:20
  • $\begingroup$ but this is not my question... $\endgroup$ Commented Sep 8, 2020 at 19:27

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