Question

If $(1-x)(1+x+x^2+x^3+x^4)$ is $\frac{31}{32}$ and $x$ is a rational number. Find the value of $1+x+x^2+x^3+x^4+x^5$.

Answer 1

Given $(1-x)(1+x+x^2+x^3+x^4)=\frac{31}{32}$......(i)
Since, $1+x+x^2+x^3+x^4=\frac{1-x^5}{1-x}$......(ii)
Substitute equation (ii) in eqaution (i), we get
$\therefore (1-x)\left(\frac{1-x^5}{1-x}\right)=\frac{31}{32}$ or $x^5=1-\frac{31}{32}=\frac{1}{32}$
or $x=\frac{1}{\sqrt[5]{532}}=\frac{1}{2}$
$\therefore 1+x+x^2+x^3+x^4+x^5=\frac{1-x^6}{1-x}=\frac{1-\frac{1}{2^6}}{1-\frac{1}{2}}=\frac{1-\frac{1}{64}}{\frac{1}{2}}$
$=2\times \frac{63}{64}=\frac{63}{32}$.

OR

$(1-x)(1+x+x^2+x^3+x^4)=\frac{31}{32}$

$\Rightarrow 1+x+x^2+x^3+x^4-x-x^2-x^3-x^4-x^5=\frac{31}{32}$

$\Rightarrow 1-x^5=\frac{31}{32}$

$\Rightarrow x^5=1-\frac{31}{32}$

$\Rightarrow x^5=\frac{1}{32}=\frac{1}{2^5}$

$\therefore x=\frac{1}{2}$

Thus,

$1+x+x^2+x^3+x^4+x^5=1+\frac{1}{2}+(\frac{1}{2}{)}^2+(\frac{1}{2}{)}^3+(\frac{1}{2}{)}^4+(\frac{1}{2}{)}^5$

$=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}$

$=\frac{63}{32}$