Question

If $\left| x \right|<\dfrac{1}{2}$, what is the value of $1+n\left[ \dfrac{x}{1-x} \right]+\left[ \dfrac{n(n+1)}{2!} \right]{{\left[ \dfrac{x}{1-x} \right]}^{2}}+....\infty $?
A. ${{\left[ \dfrac{1-x}{1-2x} \right]}^{n}}$
B. ${{(1-x)}^{n}}$
C. ${{\left[ \dfrac{1-2x}{1-x} \right]}^{n}}$
D. ${{\left( \dfrac{1}{1-x} \right)}^{n}}$

Answer 1

Hint: Here we need to find the sum of the given infinite series. By observing the series, we come to know that, the series is a binomial expansion for a rational index. So, by using the binomial theorem for the rational index, we can find the required sum. By comparing the general binomial expansion with the given expansion, we get the required values for solving it.

Formula Used: Binomial theorem for rational index:
If $n$ is a rational number, then the sum of the infinite series exists only when $\left| x \right|<1$.
If $n$ is a rational number and $\left| x \right|<1$ then the binomial theorem says
$1+nx+\dfrac{n(n-1)}{2!}{{x}^{2}}+....\infty ={{(1+x)}^{n}}$
Some of the basic expansions are:
 $1-nx+\dfrac{n(n+1)}{2!}{{x}^{2}}-\dfrac{n(n+1)(n+2)}{3!}{{x}^{3}}+....\infty ={{(1+x)}^{-n}}$
$1+nx+\dfrac{n(n+1)}{2!}{{x}^{2}}+\dfrac{n(n+1)(n+2)}{3!}{{x}^{3}}+....\infty ={{(1-x)}^{-n}}$
(Note: Here observe the signs)

Complete step by step solution: The given series is
$1+n\left[ \dfrac{x}{1-x} \right]+\left[ \dfrac{n(n+1)}{2!} \right]{{\left[ \dfrac{x}{1-x} \right]}^{2}}+....\infty $ and $\left| x \right|<\dfrac{1}{2}$
On comparing the given series with the general binomial expansion, we get the similar one as
$1+nx+\dfrac{n(n+1)}{2!}{{x}^{2}}+\dfrac{n(n+1)(n+2)}{3!}{{x}^{3}}+....\infty ={{(1-x)}^{-n}}$
Here the only difference is $x=\dfrac{x}{1-x}$
So, we can substitute $x=\dfrac{x}{1-x}$ in the general term.
Thus, we get
$1+n\left[ \dfrac{x}{1-x} \right]+\left[ \dfrac{n(n+1)}{2!} \right]{{\left[ \dfrac{x}{1-x} \right]}^{2}}+....\infty ={{(1-\dfrac{x}{1-x})}^{-n}}$
On simplifying,
$\begin{align}
  & {{(1-\dfrac{x}{1-x})}^{-n}}={{\left( \dfrac{1-x-x}{1-x} \right)}^{-n}} \\
 & \text{ }={{\left( \dfrac{1-2x}{1-x} \right)}^{-n}} \\
\end{align}$
On applying ${{\left( \dfrac{a}{b} \right)}^{-n}}={{\left( \dfrac{b}{a} \right)}^{n}}$, we get
${{\left( 1-\dfrac{x}{1-x} \right)}^{-n}}={{\left( \dfrac{1-x}{1-2x} \right)}^{n}}$
Therefore, the value of given series $1+n\left[ \dfrac{x}{1-x} \right]+\left[ \dfrac{n(n+1)}{2!} \right]{{\left[ \dfrac{x}{1-x} \right]}^{2}}+....\infty $ is ${{\left[ \dfrac{1-x}{1-2x} \right]}^{n}}$.

Option ‘A’ is correct

Note: Here, we need to remember that, while comparing the given series with the general expansions, the signs between the terms and signs of their powers in the series should be observed carefully in order to get the correct value.