Question

If $a$ satisfies the equation $a^{2017}-2a+1=0$ and $S=1+a+a^2+.....+a^{2016}$, then possible value(s) of $S$ is/ are

• A. $2016$
• B. $2018$
• C. $2017$
• D. $2$

Answer 1

The correct option is D $2$

Given $a^{2017}-2a+1=0$

$\therefore 2a-1=a^{2017}....\left(1\right)$

and $S=1+a+a^2+......+a^{2016}$

Summation of $GP$ series (finite) $=a\frac{(1-r^4)}{1-r}$

When $a=1st$ term of $GP$ series $n=$ total number of terms in the series

$r=$ common ratio between two terms

$\therefore S=\frac{1(1-a^{2017)}}{1-a}$ here $a=1$

$r=a$

$n=2017$

putting value of $a^{2017}$ from $\left(D\right)$

$S=\frac{1(1-2a+1)}{1-a}=\frac{2(1-a)}{1-a}=2$

Option $D$