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

Hint: 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.