Question

If for some positive integer $n>1$, we have $1+x+x^2+....+x^{n-1}=0$, then $\sum _{k=0}^n(x^k+\frac{1}{x^k})$ is

• A. $-2$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

The correct option is D $2$

We have $1+x+x^2+..........+x^{n-1}=0$, where $n>1$

$\to 1+x+x^2+..........+x^{n-1}$, is a geometric progression, with common ratio $x$

The sum of the geometric progression is $S_n=\frac{a_1(1-r^n)}{1-r}$, where $a_1$ is first term of the series and $r$ is the common ratio.

So sum of the given geometric progression $S_n=\frac{1(1-x^n)}{1-x}$

As $S_n=0$, Hence $\frac{1(1-x^n)}{1-x}=0$

$\to x\ne 1$ and $x^n=1$

So $x=-1$ and $n$ is some even integer.

Now the value of the sum, $S_k={\sum }_{k=0}^n$ $(x^k+\frac{1}{x^k})$ by putting the value of $x$ and $n$, will be:

$\to$ ${\sum }_{k=0}^n$ $({(-1)}^k+\frac{1}{{(-1)}^k})$

$\to {\sum }_{k=0}^n$$({(-1)}^k+{(-1)}^k)$

$\to {\sum }_{k=0}^n$$\left({2(-1)}^k\right)$

$\to {\sum }_{k=0}^n$$\left(2{(-1)}^k\right)$

$\to$${2(-1)}^0$ $+{2(-1)}^1$ $+{2(-1)}^2$ $+.....+{2(-1)}^n$

$\to 2(-1{)}^0+\left(2\right(-1{)}^1+2(-1{)}^2+2(-1{)}^2+....+2(-1{)}^n)$

$\to 2+(-2+2-2+....+2(-1{)}^n)$, $\because n$ is even, so the terms inside the bracket will get cancel with each-other.

$S_k=2$

So option $D$ is correct.