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

It is given that $\alpha$ and $\beta$ are roots of quadratic equation x² - x - 1=0, so sum of roots = $\alpha$ + 0 = 1, and product of roots =$=\alpha \beta =-1$
and ${\mathrm{p}}_{\mathrm{k}}={\mathrm{\alpha }}^{\mathrm{k}}+{\mathrm{\beta }}^{\mathrm{k}},\mathrm{k}\ge 1$
so, ${\mathrm{p}}_1=\mathrm{\alpha }+\mathrm{\beta }=1$
$$\begin{gathered} {\mathrm{p}}_2={\mathrm{\alpha }}^2+{\mathrm{\beta }}^2=(\mathrm{\alpha }+\mathrm{\beta }{)}^2-2\mathrm{\alpha \beta }=1+2=3 \\ {\mathrm{p}}_3={\mathrm{\alpha }}^3+{\mathrm{\beta }}^3=(\mathrm{\alpha }+\mathrm{\beta }{)}^3-3\mathrm{\alpha \beta }(\mathrm{\alpha }+\mathrm{\beta })=1+3=4 \\ {\mathrm{p}}_4={\mathrm{\alpha }}^4+{\mathrm{\beta }}^4=(\mathrm{\alpha }+\mathrm{\beta }{)}^4-4\mathrm{\alpha \beta }\left({\mathrm{\alpha }}^2+{\mathrm{\beta }}^2\right)-6{\mathrm{\alpha }}^2{\mathrm{\beta }}^2 \\ =1+12-6=7 \end{gathered}$$
$$\begin{gathered} \text{and }{\mathrm{p}}_5={\mathrm{\alpha }}^5+{\mathrm{\beta }}^5 \\ =(\mathrm{\alpha }+\mathrm{\beta }{)}^5-5\mathrm{\alpha \beta }\left({\mathrm{\alpha }}^3+{\mathrm{\beta }}^3\right)-10{\mathrm{\alpha }}^2{\mathrm{\beta }}^2(\mathrm{\alpha }+\mathrm{\beta }) \\ =1+20-10=11 \\ \because {\mathrm{p}}_3={\mathrm{p}}_5-{\mathrm{p}}_4=4 \\ {\mathrm{p}}_1+{\mathrm{p}}_2+{\mathrm{p}}_3+{\mathrm{p}}_4+{\mathrm{p}}_5=1+3+4+7+11=26 \\ \text{ but }{\mathrm{p}}_5\ne {\mathrm{p}}_2·{\mathrm{p}}_3 \end{gathered}$$