Question

If $(1-y)(1+2x+4x^2+8x^3+16x^4+32x^5)=(1-y^6)$ , (y $\ne$1), then a value of y/x is

• A. $1/2$
• B. $2$
• C. $25/24$
• D. $24/25$

Answer 1

The correct option is B $2$

$1+2x+4x^2+8x^3+16x^4+32x^5$ indicate the sum of a G.P. with a common ratio of 2x.
Hence
$S=1+2x+4x^2+8x^3+16x^4+32x^5$
$=\frac{1\left(\right(2x{)}^6-1)}{2x-1}$
$=\frac{64x^6-1}{2x-1}$
Hence
$(1-y)(1+2x+4x^2+8x^3+16x^4+32x^5)=(1-y^6)$
$1+2x+4x^2+8x^3+16x^4+32x^5=\frac{1-y^6}{1-y}$
$\frac{(2x{)}^6-1}{2x-1}=\frac{1-y^6}{1-y}$
$\frac{(2x{)}^6-1}{2x-1}=\frac{y^6-1}{y-1}$
Comparing x and y, gives us
$2x=y$
Or
$\frac{y}{x}=2$