Complex Numbers Part 2

Concept:The equation $x^{2}+x+1=0$ implies $x$ is a non-real cube root of unity, say $\omega$ or $\omega^{2}$, satisfying $\omega^{3}=1$ and $1+\omega+\omega^{2}=0$.Explanation:For any integer $k$, we have $\omega^{k}+\frac{1}{\omega^{k}}$ = $\omega^{k}+\omega^{-k}$.If $k$ is a multiple of $3$ (i.e., $k=3m$), then $\omega^{3m}=1$ and $\omega^{-3m}=1$, so the sum is $2$.If $k$ is not a multiple of $3$, then $\omega^{k}$ is either $\omega$ or $\omega^{2}$, and since $1+\omega+\omega^{2}=0$, we get $\omega^{k}+\omega^{2k}=-1$. Hence the sum is $-1$.Now consider the sum from $k=1$ to $25$: $\sum_{k=1}^{25}\left(x^{k}+\frac{1}{x^{k}}\right)^{4}$.For $k$ a multiple of $3$, the term is $2^{4}=16$. For all other $k$, the term is $(-1)^{4}=1$.Count multiples of $3$ between $1$ and $25$: $3,6,9,12,15,18,21,24$ — that is $8$ terms.Remaining $25-8=17$ terms are $1$.So total value $=17 \times 1 + 8 \times 16 = 17 + 128 = 145$.Answer:$145$ corresponds to option B.