Let \alpha and \beta be the roots of the equation x2 - x - 1 = 0. If pk = {\left( \alpha \right)^k} + {\left( \beta \right)^k} , k \ge 1, then which one of the following statements is not true?

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Answer: D. p5 = p2 · p3 x2 - x - 1 = 0 	herefore \alpha 2 - \alpha - 1 = 0 \Rightarrow \alpha 2 = \alpha + 1 	herefore \alpha 3 = \alpha 2 + \alpha = \alpha + 1 + \alpha = 2\alpha + 1 Now \alpha 4 = 2\alpha 2 + \alpha = 2(\alpha + 1) + \alpha = 3\alpha + 2 Now \alpha 5 = 3\alpha 2 + 2\alpha = 3(\alpha + 1) + 2\alpha = 5\alpha + 3 Given pk = {\left( \alpha ight)^k} + {\left( \beta ight)^k} 	herefore p5 = {\left( \alpha ight)^5} + {\left( \beta ight)^5} = 5\alpha + 3 + 5\beta + 3 = 5(\alpha + \beta ) + 6 = 5(1) + 6 [As \alpha + \beta = 1] = 11 Now p2 · p3 = (\alpha 2 + \beta 2).(\alpha 3 + \beta 3) = ( \alpha + 1 + \beta + 1)(2\alpha + 1 + 2\beta + 1) = ( \alpha + \beta + 2)(2(\alpha + \beta ) + 2) = (1 + 2)(2 + 2) = 12 	herefore p5 
e p2 ·p3 So option (D) is wrong.

Let $$\alpha $$ and $$\beta $$ be the roots of the equation x2 - x - 1 = 0. If pk = $${\left( \alpha \right)^k} + {\left( \beta \right)^k}$$ , k $$ \ge $$ 1, then which one of the following statements is not true?

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