• # question_answer Let $\alpha$ and $\beta$ be the roots of the equation ${{x}^{2}}-x-1=0$.  If ${{p}_{k}}={{(\alpha )}^{k}}+{{(\beta )}^{k}}$, $k\ge 1,$then which one of the following statements is not true? [JEE MAIN Held on 07-01-2020 Evening] A) ${{p}_{3}}={{p}_{5}}{{p}_{4}}$ B) $({{p}_{1}}+{{p}_{2}}+{{p}_{3}}+{{p}_{4}}+{{p}_{5}})=26$ C) ${{p}_{5}}=11$ D) ${{p}_{5}}={{p}_{2}}\cdot {{p}_{3}}$

 [d] $\because \,\,\,\,\,\alpha ,\beta$are roots of ${{x}^{2}}x1=0$ ...(i) $\therefore \,\,\,\,{{\alpha }^{2}}-\alpha -1=0$ $\Rightarrow \,\,\,\,{{\alpha }^{n+2}}-{{\alpha }^{n+1}}-{{\alpha }^{n}}=0$                                ...(ii) Similarly, ${{\beta }^{n+2}}-{{\beta }^{n+1}}-{{\beta }^{n}}=0$              ...(iii) From eq. (ii) + (iii), we get ${{\alpha }^{n+2}}+{{\beta }^{n+2}}=({{\alpha }^{n+1}}+{{\beta }^{n+1}})+({{\alpha }^{n}}+{{\beta }^{n}})$ $\therefore \,\,\,\,{{p}_{n+2}}={{p}_{n+1}}+{{p}_{n}}$ For $n=0,\,\,{{p}_{0}}={{\alpha }^{0}}+{{\beta }^{0}}=2$ For $n=1,\text{ }{{p}_{1}}=\alpha +\beta =1$ and      ${{p}_{2}}={{p}_{0}}+{{p}_{1}}=2+1=3$ ${{p}_{3}}={{p}_{2}}+{{p}_{1}}=3+1=4$ ${{p}_{4}}={{p}_{3}}+{{p}_{2}}=4+3=7$ ${{p}_{5}}={{p}_{4}}+{{p}_{3}}=7+4=11$