Animated Solution for Mathematics - Complex Numbers: If $\alpha$ is a root of the equation $x^2+x+1=0$ and ${\sum }_{k=1}^n({\alpha }^k+\frac{1}{{\alpha }^k}{)}^2=20$, then $n$ is equal to

The quadratic equation $x^2+x+1=0$ is a fundamental structure in complex number theory. Recognizing the identity $x^3-1=(x-1)(x^2+x+1)$, we observe that for $x eq 1$, the roots satisfy $x^3=1$.

These roots are denoted as $\omega$ and ${\omega }^2$. By setting $\alpha =\omega$, we establish the foundation for evaluating the given summation.

We aim to evaluate the sum:

Defining the general term as $T_k=({\omega }^k+{\omega }^{-k}{)}^2$, we utilize the property ${\omega }^{-1}={\omega }^2$ to simplify the expression:

Let us examine the sequence of terms: For $k=1$: $T_1=(\omega +{\omega }^2{)}^2=(-1{)}^2=1$. For $k=2$: $T_2=({\omega }^2+{\omega }^4{)}^2=({\omega }^2+\omega {)}^2=(-1{)}^2=1$. For $k=3$: $T_3=({\omega }^3+{\omega }^6{)}^2=(1+1{)}^2=4$.

The sequence $T_k$ follows a periodic rhythm of $1,1,4,1,1,4,\dots$ with a period of 3.

The sum of one complete cycle of three terms is:

We seek the value of $n$ such that the sum equals 20. Dividing 20 by the cycle sum of 6, we find:

This indicates 3 full cycles (covering $3\times 3=9$ terms) with a sum of 18. To reach the target of 20, we require two additional terms from the next cycle.

The 10th term is 1 (sum becomes $18+1=19$). The 11th term is 1 (sum becomes $19+1=20$).

Thus, the value of $n$ is 11.