Question

If $y=x-x^2+x^3-x^4+\dots \infty$, then the value of $x$ will be

• A. $\frac{y+1}{y}$
• B. $\frac{y}{1+y}$
• C. $\frac{y-1}{y}$
• D. $\frac{y}{1-y}$

Answer 1

The correct option is D

$\frac{y}{1-y}$

Solve the series:

Step 1: Identify series

Given that, $y=x-x^2+x^3-x^4+\dots \infty$

Clearly this series is G.P series as ratio of two consecutive terms is constant.

$\begin{array}{rcl}r& =& \frac{a_2}{a_1}\\ & =& \frac{-x^2}{x}\\ & =& -x\end{array}$

Step 2: Find the sum.

Since the series is infinite so the sum of the series is given by:

$S_{\infty }=\frac{a}{1-r};$

Where , $\mathrm{a}$is the first term

Here, $a=x,r=-x$

$$\begin{gathered} \Rightarrow y=\frac{x}{1-\left(-x\right)} \\ =\frac{x}{1+x} \end{gathered}$$

$$\begin{gathered} \Rightarrow y\left(1+x\right)=x \\ \Rightarrow y+yx=x \\ \Rightarrow y=x-yx \\ \Rightarrow y=x\left(1-y\right) \\ \Rightarrow x=\frac{y}{1-y} \end{gathered}$$

Hence, option (D) is correct.