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If $S=\frac{2^2-1}{2}+\frac{3^2-2}{6}+\frac{4^2-3}{12}+\ldots$ upto 10 terms, then $S$ is equal to
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Verified Answer
The correct answer is:
$\frac{120}{11}$
$$
\begin{aligned}
& \text { We know, } T_n=\frac{(n+1)^2-n}{n(n+1)} \\
& =1+\left(\frac{1}{n}-\frac{1}{n+1}\right) \\
& \therefore \quad S_{10}=10+\left(1-\frac{1}{11}\right)=\frac{120}{11} \\
&
\end{aligned}
$$
\begin{aligned}
& \text { We know, } T_n=\frac{(n+1)^2-n}{n(n+1)} \\
& =1+\left(\frac{1}{n}-\frac{1}{n+1}\right) \\
& \therefore \quad S_{10}=10+\left(1-\frac{1}{11}\right)=\frac{120}{11} \\
&
\end{aligned}
$$
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