1. Given equation and root properties

Since $\alpha$ is a root of $x^2+x+1=0,$ we know its roots are the non-real cube roots of unity: $\alpha^3=1, \qquad \alpha\ne 1.$ Also, $1+\alpha+\alpha^2=0.$

Because $\alpha^3=1$, we also have $\frac{1}{\alpha}=\alpha^2.$ So, $\alpha^k+\frac{1}{\alpha^k}=\alpha^k+\alpha^{-k}.$

2. Evaluate the repeating values

Let $T_k=\left(\alpha^k+\frac{1}{\alpha^k}\right)^2.$ Since $\alpha^3=1$, powers repeat modulo $3$.

We compute according to $k \pmod 3$:

• If $k\equiv 0 \pmod 3$, then $\alpha^k=1$, so $\alpha^k+\frac{1}{\alpha^k}=1+1=2,$ hence $T_k=2^2=4.$

• If $k\equiv 1 \pmod 3$, then $\alpha^k+\frac{1}{\alpha^k}=\alpha+\alpha^2=-1,$ so $T_k=(-1)^2=1.$

• If $k\equiv 2 \pmod 3$, then $\alpha^k+\frac{1}{\alpha^k}=\alpha^2+\alpha=-1,$ so again $T_k=1.$

Thus the sequence $T_k$ is periodic with pattern: $1,1,4,1,1,4,\dots$

3. Sum one full block

For every block of $3$ consecutive terms, $1+1+4=6.$

Let $n=3q+r, \qquad r\in\{0,1,2\}.$ Then $\sum_{k=1}^{n} T_k = 6q + \text{(sum of first $r$ terms of pattern)}.$

The partial sums for remainder are:

• $r=0$: extra sum $=0$
• $r=1$: extra sum $=1$
• $r=2$: extra sum $=1+1=2$

So possible sums are: $6q, \quad 6q+1, \quad 6q+2.$

We are given $\sum_{k=1}^{n}\left(\alpha^k+\frac{1}{\alpha^k}\right)^2=20.$ So we solve: $6q + \text{remainder part} = 20.$

Now, $20=18+2=6\cdot 3+2.$ Hence, $q=3, \qquad r=2.$ Therefore, $n=3q+r=3\cdot 3+2=11.$

4. Final answer

$\boxed{11}$

5. Comparison with stored correct answer

Stored correct answer = $11$.

This matches the derived answer.