solve Q 19

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+... \mathrm{upto} 11 \mathrm{terms} \\ \mathrm{Numerator} \mathrm{follows} \mathrm{an} \mathrm{A}.\mathrm{P}. \mathrm{with} \mathrm{first} \mathrm{term} 3 \mathrm{and} \mathrm{common} \mathrm{difference} 2, \mathrm{hence} \\ {\mathrm{T}}_{\mathrm{n}}=\frac{3+\left(\mathrm{n}-1\right)2}{1^2+2^2+3^2+....+{\mathrm{n}}^2} \\ =\frac{2\mathrm{n}+1}{{\displaystyle \frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}}} \\ =6 \left[\frac{1}{\mathrm{n}\left(\mathrm{n}+1\right)}\right] \\ {\mathrm{T}}_{\mathrm{n}}=6\left[\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+1}\right] \\ {\mathrm{S}}_{11}={\mathrm{T}}_1+{\mathrm{T}}_2+{\mathrm{T}}_3+...+{\mathrm{T}}_{11} \\ =6\left[\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\right] \\ =6\left[1-\frac{1}{12}\right] \\ =6\left[\frac{11}{12}\right] \\ =\frac{11}{6} \left[\mathrm{Answer}\right] \end{gathered}$$

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