Question

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

• A. $\frac{1}{2}$
• B. 2
• C. $\frac{1}{4}$
• D. 4

Answer 1

Answer: (B)

By the given equation, we have
$\left(1-p\right)\left[\frac{1-{\left(2x\right)}^6}{1-2x}\right]=1-p^6$
$\Rightarrow \frac{1-{\left(2x\right)}^6}{1-2x}=\frac{1-p^6}{1-p}$
$\Rightarrow p=2x$
$\Rightarrow \frac{p}{x}=2$