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If $$\sum_{r=1}^nt_r=\frac{n(n+1)(n+2)}{12}$$, then value of $$\sum_{r=1}^n\frac{1}{t_r}$$ is

Now, how can we find the series (i.e. formulae of $n^{th}$ term) from the sum? The answer is $\frac{4n}{n+1}$

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  • $\begingroup$ the index is $r$ but you have $t_n$? $\endgroup$ – user715522 Feb 25 at 10:50
  • $\begingroup$ The posted solutions appear to assume that you wanted the initial sum to hold $\forall n$ but of course you don't say that anywhere. If you are making that assumption, you should make it explicitly. $\endgroup$ – lulu Feb 25 at 10:54
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$$t_n=\frac{n(n+1)(n+2)}{4}-\frac{(n-1)n(n+1)}{4}=\frac{n(n+1)}{4}.$$ Now, use $$\frac{4}{n(n+1)}=4\left(\frac{1}{n}-\frac{1}{n+1}\right)$$ and a telescopic summation.

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  • $\begingroup$ @Usercomingsoon Thank you for your accepting. Now, show please your attempts, otherwise this topic would be deleted. $\endgroup$ – Michael Rozenberg Feb 25 at 11:12
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Observe that $$ t_n=\frac{n(n+1)(n+2)}{12}-\frac{(n-1)n(n+1)}{12}=\frac{n(n+1)}{4} $$ and $$ \frac{1}{t_n}=\frac{4}{n(n+1)}=\frac{4}{n}-\frac{4}{n+1}, $$ and hence $$ \sum_{k=1}^n\frac{1}{t_k}=\left(\frac{4}{1}-\frac{4}{2}\right)+\left(\frac{4}{2}-\frac{4}{3}\right)+\cdots+ \left(\frac{4}{n-1}-\frac{4}{n}\right)+\left(\frac{4}{n}-\frac{4}{n+1}\right)\\ =\frac{4}{1}-\frac{4}{n+1}=\frac{4n}{n+1} $$

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The sum and difference are inverse operations:

$\begin{align*} S_n &= \sum_{1 \le k\le n} t_k \\ t_n &= S_ n - S_{n - 1} \end{align*}$

Thus you can get the term $t_n$ knowing the sum, and set up the sum of reciprocals.

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