Question

If $\left( {1 - y} \right)\left( {1 + 2x + 4{x^2} + 8{x^3} + 16{x^4} + 32{x^5}} \right) = \left( {1 - {y^6}} \right)$, $\left( {y \ne 1} \right)$, then a value of $\dfrac{y}{x}$ is
A.$\dfrac{1}{2}$
B.2
C.$\dfrac{{25}}{{24}}$
D.$\dfrac{{24}}{{25}}$

Answer 1

Here, we will first find the sum of the geometric series $1 + 2x + 4{x^2} + 8{x^3} + 16{x^4} + 32{x^5}$ using the formula to calculate the sum of first $n$ terms of a geometric progression, $S = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$. Then we will use the obtained value of $1 + 2x + 4{x^2} + 8{x^3} + 16{x^4} + 32{x^5}$ in the given equation and simplify it to find the required value.

Complete step-by-step answer:
We are given that the equation
$\left( {1 - y} \right)\left( {1 + 2x + 4{x^2} + 8{x^3} + 16{x^4} + 32{x^5}} \right) = \left( {1 - {y^6}} \right){\text{ ......eq.(1)}}$
We know that a geometric progression is a sequence of numbers in which each is multiplied by the same factor to obtain the next number in the sequence.
Since $1 + 2x + 4{x^2} + 8{x^3} + 16{x^4} + 32{x^5}$ is the sum of a geometric progression with a common ratio $2x$.
Hence, we have the sum is $S = 1 + 2x + 4{x^2} + 8{x^3} + 16{x^4} + 32{x^5}$.
We know that the formula to calculate the sum of first $n$ terms of a geometric progression is $S = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$.
First, we will find the number of terms $n$ and the first term $a$ in the above series $S$.
$n = 6$
$a = 1$
$r = 2x$
Using the above values of $n$, $r$ and $a$ in the above formula of sum of geometric progression.

$$\Rightarrow S = \dfrac{{1\left( {{{\left( {2x} \right)}^6} - 1} \right)}}{{2x - 1}} \\\ \Rightarrow S = \dfrac{{64{x^6} - 1}}{{2x - 1}} \\\$$

Thus, we have $1 + 2x + 4{x^2} + 8{x^3} + 16{x^4} + 32{x^5} = \dfrac{{64{x^6} - 1}}{{2x - 1}}$.
Dividing the equation ${\text{(1)}}$ by $1 - y$ on both sides, we get

$$\Rightarrow \dfrac{{\left( {1 - y} \right)\left( {1 + 2x + 4{x^2} + 8{x^3} + 16{x^4} + 32{x^5}} \right)}}{{1 - y}} = \dfrac{{\left( {1 - {y^6}} \right)}}{{1 - y}} \\\ \Rightarrow 1 + 2x + 4{x^2} + 8{x^3} + 16{x^4} + 32{x^5} = \dfrac{{\left( {1 - {y^6}} \right)}}{{1 - y}} \\\$$

Using the value of $1 + 2x + 4{x^2} + 8{x^3} + 16{x^4} + 32{x^5}$ in the above equation, we get

$$\Rightarrow \dfrac{{64{x^6} - 1}}{{2x - 1}} = \dfrac{{1 - {y^6}}}{{1 - y}} \\\ \Rightarrow \dfrac{{64{x^6} - 1}}{{2x - 1}} = \dfrac{{{y^6} - 1}}{{y - 1}} \\\$$

Comparing $x$ and $y$ in the above equation, we get
$\Rightarrow 2x = y$
Dividing the above equation by $x$ on both sides, we get

$$\Rightarrow \dfrac{{2x}}{x} = \dfrac{y}{x} \\\ \Rightarrow 2 = \dfrac{y}{x} \\\ \Rightarrow \dfrac{y}{x} = 2 \\\$$

Hence, option B is correct.

Note: While solving this question, be careful when we find the sum of the geometric series using the formula of first $n$ terms of a geometric progression, $S = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$. This is the key point of the question, some students try to solve this equation directly and end up with a long solution, which is time consuming and mostly leads to wrong answers.