Question

The sum of the series $\frac{3}{4}+\frac{5}{36}+\frac{7}{144}+...............$ upto $11$ terms is

• A. $\frac{120}{121}$
• B. $\frac{143}{144}$
• C. $1$
• D. $\frac{144}{143}$

Answer 1

The correct option is B

$\frac{143}{144}$

Explanation for the correct option

Given: $\frac{3}{4}+\frac{5}{36}+\frac{7}{144}+...............$

Let $S=\frac{3}{4}+\frac{5}{36}+\frac{7}{144}+...............$

$$\begin{gathered} \frac{3}{4}=\frac{2^2-1^2}{2^2·1^2} \\ \frac{5}{36}=\frac{3^2-2^2}{3^2·2^2} \\ \frac{7}{144}=\frac{4^2-3^2}{4^2·3^2} \end{gathered}$$

………

similarly, ${11}^{th}$ term$=\frac{{12}^2-{11}^2}{{11}^2·{12}^2}$

$$\begin{gathered} \therefore \frac{3}{4}+\frac{5}{36}+\frac{7}{144}+...............=\frac{2^2-1^2}{2^2·1^2}+\frac{3^2-2^2}{3^2·2^2}+.........\frac{{12}^2-{11}^2}{{12}^2·{11}^2} \\ \Rightarrow S=\frac{2^2}{2^2·1^2}-\frac{1^2}{2^2·1^2}+\frac{3^2}{3^2·2^2}-\frac{2^2}{3^2·2^2}+...........+\frac{{12}^2}{{12}^2·{11}^2}-\frac{{11}^2}{{12}^2·{11}^2} \\ \Rightarrow S=\frac{1}{1}-\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{3^2}+.................+\frac{1}{{11}^2}-\frac{1}{{12}^2} \\ \Rightarrow S=1-\frac{1}{{12}^2} \\ \Rightarrow S=1-\frac{1}{144} \\ \Rightarrow S=\frac{143}{144} \end{gathered}$$

Hence, $\mathrm{Option}\left(\mathrm{B}\right)$ is correct.