1. Tardigrade
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  3. Mathematics
  4. The sum to n terms of (3/12 ⋅ 22)+(5/22 ⋅ 32)+(7/32 ⋅ 42)+⋅s is (624/625). The value of n is

Q. The sum to $n$ terms of $\frac{3}{1^{2} \cdot 2^{2}}+\frac{5}{2^{2} \cdot 3^{2}}+\frac{7}{3^{2} \cdot 4^{2}}+\cdots$ is $\frac{624}{625}$. The value of $n$ is

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Solution:

$n^{t h}$ term $=\frac{2 n+1}{n^{2}(n+1)^{2}}=\frac{(n+1)^{2}-n^{2}}{n^{2}(n+1)^{2}}=\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}$.
The sum to $n$ terms is
$=\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)+\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)+\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right)+\ldots+\left(\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}\right) $
$=1-\frac{1}{(n+1)^{2}} . $
$\therefore 1-\frac{1}{(n+1)^{2}}=\frac{624}{625}=1-\frac{1}{625} $
$\Rightarrow n+1=25$
$ \Rightarrow n=24 $

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