Question

What is the sum of the series $1+\dfrac{1}{8}+\dfrac{1\cdot 3}{8\cdot 16}+\dfrac{1\cdot 3\cdot 5}{8\cdot 16\cdot 24}+...+\infty $?
A. $\dfrac{2}{\sqrt{3}}$
B. $2\sqrt{3}$
C. $\dfrac{\sqrt{3}}{2}$
D. $\dfrac{1}{2\sqrt{3}}$

Answer 1

Hint: In this question, we are to find the sum of the given series. Here we apply the binomial theorem for the rational index to find the sum of the given series. This is because given series has rational numbers.

Formula used: The binomial theorem for rational index:
$1+nx+\dfrac{n(n-1)}{2!}{{x}^{2}}+\dfrac{n(n-1)(n-2)}{3!}{{x}^{3}}+...={{(1+x)}^{n}}$
Where $\left| x \right|<1$
We can calculate the factorial of a number as
$\begin{align}
  & n!=n\cdot (n-1)\cdot (n-2)....(n-r+1) \\
 & \text{ }=n(n-1)! \\
\end{align}$

Complete step by step solution: Given series is
$1+\dfrac{1}{8}+\dfrac{1\cdot 3}{8\cdot 16}+\dfrac{1\cdot 3\cdot 5}{8\cdot 16\cdot 24}+...+\infty \text{ }...(1)$
Here, the numerators and the denominators are in arithmetic progression.
But we cannot find this by using arithmetic progression.
So, on applying the binomial theorem for rational index. I.e.,
$1+nx+\dfrac{n(n-1)}{2!}{{x}^{2}}+\dfrac{n(n-1)(n-2)}{3!}{{x}^{3}}+...={{(1+x)}^{n}}\text{ }...(2)$
On comparing (1) and (2), we get
$nx=\dfrac{1}{8}\text{ }...(3)$; and
$\dfrac{n(n-1)}{2!}{{x}^{2}}=\dfrac{1\cdot 3}{8\cdot 16}\text{ }...(4)$;
Squaring on both sides of (3), we get
$\begin{align}
  & nx=\dfrac{1}{8} \\
 & \Rightarrow {{n}^{2}}{{x}^{2}}=\dfrac{1}{8}\cdot \dfrac{1}{8}\text{ }...(5) \\
\end{align}$
On dividing (4) and (5), we get
$\begin{align}
  & \dfrac{\dfrac{n(n-1)}{2!}{{x}^{2}}}{{{n}^{2}}{{x}^{2}}}=\dfrac{\dfrac{1\cdot 3}{8\cdot 16}}{\dfrac{1}{8}\cdot \dfrac{1}{8}} \\
 & \Rightarrow \dfrac{n-1}{2n}=\dfrac{3}{2} \\
 & \Rightarrow n-1=3n \\
 & \Rightarrow 2n=-1 \\
 & \therefore n=\dfrac{-1}{2} \\
\end{align}$
Thus, substituting the obtained value in (3), we get
$\begin{align}
  & nx=\dfrac{1}{8} \\
 & \Rightarrow \left( \dfrac{-1}{2} \right)x=\dfrac{1}{8} \\
 & \Rightarrow x=\dfrac{-1}{4} \\
\end{align}$
Thus, the required sum is
$\begin{align}
  & 1+\dfrac{1}{8}+\dfrac{1\cdot 3}{8\cdot 16}+\dfrac{1\cdot 3\cdot 5}{8\cdot 16\cdot 24}+...+\infty ={{(1+x)}^{n}} \\
 & \Rightarrow {{S}_{\infty }}={{(1-\dfrac{1}{4})}^{\dfrac{-1}{2}}}={{\left( \dfrac{3}{4} \right)}^{\dfrac{-1}{2}}} \\
\end{align}$
Rewriting the obtained value and simplifying, we get
${{S}_{\infty }}={{\left( \dfrac{4}{3} \right)}^{\dfrac{1}{2}}}=\sqrt{\dfrac{4}{3}}=\dfrac{2}{\sqrt{3}}$
Therefore, the sum of the series $1+\dfrac{1}{8}+\dfrac{1\cdot 3}{8\cdot 16}+\dfrac{1\cdot 3\cdot 5}{8\cdot 16\cdot 24}+...+\infty $ is $\dfrac{2}{\sqrt{3}}$.

Thus, Option (A) is correct.

Note: Here in this question, we need to use the binomial theorem for rational numbers. Otherwise, we do not get the infinite sum of the series since the \[n^{th}\] term is also a rational number and it is less than $1$. By comparing the terms of this series with the terms in the binomial expansion, we get the values of $n$ and $x$. By using them, we can find the required sum.