If alpha$\alpha$ is a root of the equation x^2 + x + 1 = 0$x^{2} + x + 1 = 0$ and sum_k=1^nleft(alpha^k + frac1alpha^kright)^2 = 20$\sum_{k=1}^{n}\left(\alpha^{k} + \frac{1}{\alpha^{k}}\right)^{2} = 20$, then n$n$ is equal to

### Core Logic The equation x^2 + x + 1 = 0$x^2 + x + 1 = 0$ has complex roots which are the non-real cube roots of unity. Thus, we can set alpha = omega$\alpha = \omega$ (where omega^3 = 1$\omega^3 = 1$ and 1 + omega + omega^2 = 0$1 + \omega + \omega^2 = 0$). Let's analyze the general term block T_k = left(omega^k + frac1omega^kright)^2$T_k = \left(\omega^k + \frac{1}{\omega^k}\right)^2$: T_k = left(omega^k + omega^-kright)^2 = omega^2k + omega^-2k + 2 = omega^2k + omega^k + 2$$T_k = \left(\omega^k + \omega^{-k}\right)^2 = \omega^{2k} + \omega^{-2k} + 2 = \omega^{2k} + \omega^k + 2$$ Because omega^k$\omega^k$ is periodic with period 3$3$, let's examine the values of T_k$T_k$ for different values of k$k$: - If k$k$ is a multiple of 3$3$ (k=3m$k=3m$): omega^2k = 1, omega^k = 1 implies T_k = 1 + 1 + 2 = 4$\omega^{2k} = 1, \omega^k = 1 \implies T_k = 1 + 1 + 2 = 4$. - If k$k$ is not a multiple of 3$3$ (k=3m+1$k=3m+1$ or 3m+2$3m+2$): omega^2k + omega^k = -1 implies T_k = -1 + 2 = 1$\omega^{2k} + \omega^k = -1 \implies T_k = -1 + 2 = 1$. ### Step 1: Evaluating periodic blocks Every block of three consecutive terms (k = 1, 2, 3$k = 1, 2, 3$) contributes exactly: textSum of a block = 1 + 1 + 4 = 6$$\text{Sum of a block} = 1 + 1 + 4 = 6$$ We want the total summation to equal 20$20$. Let's divide 20$20$ by our block value 6$6$: 20 = 3 times 6 + 2$$20 = 3 \times 6 + 2$$ This means the sum must consist of 3$3$ full periodic blocks plus additional terms that add up to 2$2$. ### Step 2: Determining the final term count n The number of terms in 3$3$ full blocks is 3 times 3 = 9$3 \times 3 = 9$ terms, giving a sum of 18$18$. To get the remaining value of 2$2$, we look at the next terms: - Term 10 (k=10$k=10$, not a multiple of 3) adds 1 implies textTotal = 18 + 1 = 19$1 \implies \text{Total} = 18 + 1 = 19$. - Term 11 (k=11$k=11$, not a multiple of 3) adds 1 implies textTotal = 19 + 1 = 20$1 \implies \text{Total} = 19 + 1 = 20$. Hence, the series terminates exactly at n = 11$n = 11$. ### Pattern Recognition Whenever complex roots of unity or cyclic properties show up inside series sums, group terms into blocks based on the underlying period length (3$3$ here) to convert large sums into simple modular arithmetic arithmetic calculations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Complex Numbers Class 11 Mathematics: Sequences and Series