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The sum $\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+\ldots$. upto 11-terms is:
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Verified Answer
The correct answer is:
$\frac{11}{2}$
$\frac{11}{2}$
Given sum is
$$
\begin{aligned}
& \frac{3}{12}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+\ldots . . \\
& n \text {th term }=\mathrm{T}_n \\
& =\frac{2 n+1}{\frac{n(n+1)(2 n+1)}{6}}=\frac{6}{n(n+1)} \\
& \text { or } \mathrm{T}_n=6\left[\frac{1}{n}-\frac{1}{n+1}\right] \\
& \therefore \quad \mathrm{S}_n= \\
& \sum \mathrm{T}_n=6 \sum \frac{1}{n}-6 \sum \frac{1}{n+1}=\frac{6 n}{n}-\frac{6}{n+1} \\
& =6-\frac{6}{n+1}=\frac{6 n}{n+1} \\
& S_{11}=\frac{6 \times 11}{11+1}=\frac{66}{12}=\frac{33}{6}=\frac{11}{2} \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \frac{3}{12}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+\ldots . . \\
& n \text {th term }=\mathrm{T}_n \\
& =\frac{2 n+1}{\frac{n(n+1)(2 n+1)}{6}}=\frac{6}{n(n+1)} \\
& \text { or } \mathrm{T}_n=6\left[\frac{1}{n}-\frac{1}{n+1}\right] \\
& \therefore \quad \mathrm{S}_n= \\
& \sum \mathrm{T}_n=6 \sum \frac{1}{n}-6 \sum \frac{1}{n+1}=\frac{6 n}{n}-\frac{6}{n+1} \\
& =6-\frac{6}{n+1}=\frac{6 n}{n+1} \\
& S_{11}=\frac{6 \times 11}{11+1}=\frac{66}{12}=\frac{33}{6}=\frac{11}{2} \\
&
\end{aligned}
$$
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