Solve the following

Question:

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 ?

  1. (1) $p_{3}=p_{5}-p_{4}$

  2. (2) $P_{5}=11$

  3. (3) $\left(p_{1}+p_{2}+p_{3}+p_{4}+p_{5}\right)=26$

  4. (4) $p_{5}=p_{2} \cdot p_{3}$

Correct Option: , 4

Solution:

$\alpha^{5}=5 \alpha+3$

$\beta^{5}=5 \beta+3$

$p_{5}=5(\alpha+\beta)+6=5(1)+6$

$\left[\because\right.$ from $\left.x^{2}-x-1=0, \alpha+\beta=\frac{-b}{a}=1\right]$

$p_{5}=11$ and $p_{5}=\alpha^{2}+\beta^{2}=\alpha+1+\beta+1$

$p_{2}=3$ and $p_{3}=\alpha^{3}+\beta^{3}=2 \alpha+1+2 \beta+1$

$=2(1)+2=4$

$p_{2} \times p_{3}=12$ and $p_{5}=11 \Rightarrow p_{5} \neq p_{2} \times p_{3}$

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