Question

Sum of infinite series $1+\frac{2}{3}·\frac{1}{2}+\frac{2}{3}·\frac{5}{6}·\frac{1}{2^2}+\frac{2}{3}·\frac{5}{6}·\frac{8}{9}·\frac{1}{2^3}+...$ is:

• A. $2^{\frac{1}{3}}$
• B. $4^{\frac{1}{3}}$
• C. $8^{\frac{1}{3}}$
• D. $2^{\frac{1}{5}}$

Answer 1

The correct option is B

$4^{\frac{1}{3}}$

Explanation for correct option

Calculating the sum of the given infinite series:

Given series is $1+\frac{2}{3}·\frac{1}{2}+\frac{2}{3}·\frac{5}{6}·\frac{1}{2^2}+\frac{2}{3}·\frac{5}{6}·\frac{8}{9}·\frac{1}{2^3}+...$

We know that,

${(1+x)}^n=1+nx+\frac{n(n-1)}{2!}{x}^2+\frac{n(n-1)(n-2)}{3!}{x}^3+...$

By comparing the two equations we get:

$nx=\frac{2}{3}·\frac{1}{2}$ and $\frac{n(n-1)}{2!}{x}^2=\frac{2}{3}·\frac{5}{6}·\frac{1}{2^2}$.

$\Rightarrow n^2{x}^2=\frac{4}{9}·\frac{1}{4};\frac{n(n-1)}{2!}{x}^2=\frac{2}{3}·\frac{5}{6}·\frac{1}{2^2}$

Dividing the two equations:

$$\begin{gathered} \Rightarrow \frac{{\displaystyle \frac{n(n-1)}{2!}}{x}^2}{n^2{x}^2}=\frac{{\displaystyle \frac{2}{3}}·{\displaystyle \frac{5}{6}}·{\displaystyle \frac{1}{2^2}}}{{\displaystyle \frac{4}{9}}·{\displaystyle \frac{1}{4}}} \\ \Rightarrow \frac{n-1}{2n}=\frac{5}{4} \\ \Rightarrow \frac{n-1}{n}=\frac{5}{2} \\ \Rightarrow 1-\frac{1}{n}=\frac{5}{2} \\ \Rightarrow \frac{1}{n}=1-\frac{5}{2} \\ \Rightarrow \frac{1}{n}=\frac{-3}{2} \\ \Rightarrow n=\frac{-2}{3} \end{gathered}$$

Substitute value of $n$ in $nx=\frac{2}{3}·\frac{1}{2}$:

$$\begin{gathered} \Rightarrow \left(\frac{-2}{3}\right)x=\frac{2}{3}·\frac{1}{2} \\ \Rightarrow x=\frac{2}{3}·\frac{1}{2}·\left(\frac{-3}{2}\right) \\ \Rightarrow x=\frac{-1}{2} \end{gathered}$$

Therefore, sum of the series will be:

$$\begin{gathered} =1+\frac{2}{3}·\frac{1}{2}+\frac{2}{3}·\frac{5}{6}·\frac{1}{2^2}+\frac{2}{3}·\frac{5}{6}·\frac{8}{9}·\frac{1}{2^3}+... \\ ={\left(1-\frac{1}{2}\right)}^{\frac{-2}{3}} \\ ={\left(\frac{1}{2}\right)}^{\frac{-2}{3}} \\ =2^{\frac{2}{3}} \\ =4^{\frac{1}{3}} \end{gathered}$$

Thus, the sum of infinite series is $4^{1/3}$.

Hence, the correct answer is Option (B).