Question

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

• A. $(p_1+p_2+p_3+p_4+p_5)=26$
• B. $p_5=11$
• C. $p_5=p_2\cdot p_3$
• D. $p_3=p_5-p_4$

Answer 1

The correct option is C

$p_5=p_2\cdot p_3$

Explanation for the correct option:

Step 1: Determining the values

For $x^2-x-1=0$

Sum of roots $\alpha +\beta =1$

Product of roots $\alpha \beta =-1$

Since, $\alpha \text{and}\beta$ are the roots of the given equation

$$\begin{gathered} {\alpha }^2-\alpha -1=0 \\ {\alpha }^2=\alpha +1\dots \dots \left(1\right) \\ \text{Similarly} \\ {\beta }^2-\beta -1=0 \\ {\beta }^2=\beta +1\dots \dots \left(2\right) \end{gathered}$$

$$\begin{gathered} \because p_k={\alpha }^k+{\beta }^k(k\ge 1) \\ \Rightarrow P_k={\alpha }^{k-2}\left({\alpha }^2\right)+{\beta }^{k-2}({\beta }^2) \\ ={\alpha }^{k-2}(\alpha +1)+{\beta }^{k-2}(\beta +1)\left[\text{From equations}\left(1\right)\text{and}\left(2\right)\right] \\ ={\alpha }^{k-1}+{\alpha }^{k-2}+{\beta }^{k-1}+{\beta }^{k-2} \\ =p_{k-1}+p_{k-2}...\left(3\right)\left[\because p_k={\alpha }^k+{\beta }^k\right] \end{gathered}$$

Value of $p_1={\alpha }^1+{\beta }^1$

$p_1=1\mathbf{}\left[\because \alpha +\beta =1\right]$

Value of $p_2={\alpha }^2+{\beta }^2$

$p_2={\left(\alpha +\beta \right)}^2-2\alpha \beta \mathbf{}\left[\because a^2+b^2={(a+b)}^2-2\mathrm{ab}\right]$

$p_2={\left(1\right)}^2-2(-1)\left[\because (\alpha +\beta )=1,\mathrm{\alpha \beta }=-1\right]$

$p_2=1+2$

$p_2=3$

Value of $p_3=p_2+p_1[\mathrm{From}\mathrm{equation}3]$

$$\begin{gathered} \therefore p_3=1+3 \\ =4 \end{gathered}$$

Value of $p_4=p_3+p_2[\mathrm{From}\mathrm{equation}3]$

$$\begin{gathered} \therefore p_4=4+3 \\ =7 \end{gathered}$$

Value of $p_5=p_4+p_3[\mathrm{From}\mathrm{equation}3]$

$$\begin{gathered} \therefore p_5=7+4 \\ =11 \end{gathered}$$

Option (C)

$$\begin{gathered} LHS \\ {p}_5=11 \\ RHS \\ {p}_3{p}_2=4\times 3 \\ LHS\ne RHS \end{gathered}$$

Thus , statement (C) is false

Explanation for the in-correct option:

Option (A)

$$\begin{gathered} \because p_1+p_2+p_3+p_4+p_5=1+3+4+7+11 \\ =26 \end{gathered}$$

Thus, the statement (A) is true

Option (B)

From above we know that, $p_5=11$

Thus, statement (B) is true

Option (D)

$p_5=p_4+p_3[\mathrm{From}\mathrm{equation}3]$

$p_3=p_5-p_4$

Thus, the statement (D) is true.
Hence, the correct answer is option (C)