Question

Assertion :If $z^2+z+1=0$ and $n$ is a natural number, then $\sum _{k=1}^n(z^k+z^{-k}{)}^2=n+3\left[\frac{n}{3}\right]$ where $\left[x\right]$ denotes the greatest integer $\le x$ Reason: If $w\ne 1$ is a cube root of unity, then $w^k+w^{-k}=\{\begin{array}{cc}0-1& \text{if k is not a multiple of 3}\\ 2& \text{if k is a multiple of 3}\end{array}$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

$z^2+z+1=0$ then $z=\omega$ and ${\omega }^2$
We know that

${\omega }^3=1$ also ${\omega }^2+\omega =-1$

Now,

When $k$ divided by $3$ leaves remainder $1$

i.e. $k=3n+1$,so $w^{-k}={\omega }^2$ and ${\omega }^k=\omega$
then ${\omega }^k+{\omega }^{-k}=-1$
When $k$ divided by $3$ leaves remainder $2$

i.e. $k=3n+2$, so ${\omega }^{-k}=\omega$ and ${\omega }^k={\omega }^2$
then ${\omega }^k+{\omega }^{-k}=-1$

When $k$ divided by $3$ leaves remainder $0$

i.e. $k=3n$, so ${\omega }^{-k}=1$ and ${\omega }^k=1$
then ${\omega }^k+{\omega }^{-k}=2$

Thus, considering $\omega$ we have
$\sum _1^3{(z^k+z^{-k})}^2=1+1+4=6$

which can be written as $3+3\left[\frac{3}{3}\right]$
$\sum _1^6{(z^k+z^{-k})}^2=1+1+4+1+1+4=12$

which can be written as $6+3\left[\frac{6}{3}\right]$
$\sum _1^9{(z^k+z^{-k})}^2=1+1+4+1+1+4+1+1+4=18$

which can be written as $9+3\left[\frac{9}{3}\right]$
and so on...
Thus, we can say that
$\sum _1^n{(z^k+z^{-k})}^2=n+3\left[\frac{n}{3}\right]$

Hence, option A.