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Q.
If $(1-p)(1+2x+4x^2+8x^3+16x^4+32x^5)=1-p^6,p \neq1$ , then a value of $\frac{p}{x}$ is
Sequences and Series
Solution:
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 $