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(D) Given the equation $x^2-x-1=0$,the roots $\alpha$ and $\beta$ satisfy $\alpha+\beta=1$ and $\alpha\beta=-1$.Since $\alpha$ and $\beta$ are roots,$

Vedclass · Accepted Answer

$p_5=p_2 \cdot p_3$

Vedclass · Answer

(D) Given the equation $x^2-x-1=0$,the roots $\alpha$ and $\beta$ satisfy $\alpha+\beta=1$ and $\alpha\beta=-1$.Since $\alpha$ and $\beta$ are roots,$\alpha^k-\alpha^{k-1}-\alpha^{k-2}=0$ and $\beta^k-\beta^{k-1}-\beta^{k-2}=0$.Adding these gives the recurrence relation $p_k=p_{k-1}+p_{k-2}$.$p_1 = \alpha+\beta = 1$.$p_2 = (\alpha+\beta)^2 - 2\alpha\beta = 1^2 - 2(-1) = 3$.$p_3 = p_2+p_1 = 3+1 = 4$.$p_4 = p_3+p_2 = 4+3 = 7$.$p_5 = p_4+p_3 = 7+4 = 11$.Now check the options:$A: p_1+p_2+p_3+p_4+p_5 = 1+3+4+7+11 = 26$ (True).$B: p_5 = 11$ (True).$C: p_5-p_4 = 11-7 = 4 = p_3$ (True).$D: p_2 \cdot p_3 = 3 \cdot 4 = 12 
eq p_5$ (False).Thus,the statement that is not true is $p_5=p_2 \cdot p_3$.

Let $\alpha$ and $\beta$ be the roots of the equation $x^2-x-1=0$. If $p_{k}=(\alpha)^{k}+(\beta)^{k}, k \geq 1$,then which one of the following statements is not true?

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