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(D) Given the equation $x^{2}-x-1=0$,by Vieta's formulas,$\alpha+\beta=1$ and $\alpha\beta=-1$.Since $\alpha$ and $\beta$ are roots,$\alpha^{2}-\alpha

Vedclass · Accepted Answer

$p_{5}=p_{2} \cdot p_{3}$

Vedclass · Answer

(D) Given the equation $x^{2}-x-1=0$,by Vieta's formulas,$\alpha+\beta=1$ and $\alpha\beta=-1$.Since $\alpha$ and $\beta$ are roots,$\alpha^{2}-\alpha-1=0$ and $\beta^{2}-\beta-1=0$.Multiplying by $\alpha^{k-2}$ and $\beta^{k-2}$ respectively,we get $\alpha^{k}-\alpha^{k-1}-\alpha^{k-2}=0$ and $\beta^{k}-\beta^{k-1}-\beta^{k-2}=0$.Adding these,we obtain the recurrence relation $p_{k}=p_{k-1}+p_{k-2}$.Calculating the values:$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$Checking the options:$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 $D$.

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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