Question

If x² – x + 1 = 0, then the value of $\sum_{n = 1}^{5}\left( x^{n} + \frac{1}{x^{n}} \right)^{2}$is -$\sum_{n = 1}^{5}\left( x^{n} + \frac{1}{x^{n}} \right)^{2}$is –

• A. 8
• B. 10
• C. 12
• D. None of these

Answer 1

Answer: (A)

8

Answer 2

Sol. x² – x + 1 = 0

Ž x = $\frac{1 \pm \sqrt{3}i}{2}$ = –w, –w²

\ $\sum_{n = 1}^{5}\left( x^{2n} + \frac{1}{x^{2n}} + 2 \right)$

Ž $\left( x^{2} + \frac{1}{x^{2}} + 2 \right)$ + $\left( x^{4} + \frac{1}{x^{4}} + 2 \right)$ + $\left( x^{6} + \frac{1}{x^{6}} + 2 \right)$

+ $\left( x^{8} + \frac{1}{x^{8}} + 2 \right)$ + $\left( x^{10} + \frac{1}{x^{10}} + 2 \right)$

Ž (w² + w⁴ + w⁶ + w⁸ + w¹⁰) +

$\left( \frac{1}{\omega^{2}} + \frac{1}{\omega^{4}} + \frac{1}{\omega^{6}} + \frac{1}{\omega^{8}} + \frac{1}{\omega^{10}} \right)$ + 10

Ž – 1 – 1 + 10 = 8.

Hence (1) is correct answer.