A) \[{{p}_{3}}={{p}_{5}}{{p}_{4}}\]
B) \[({{p}_{1}}+{{p}_{2}}+{{p}_{3}}+{{p}_{4}}+{{p}_{5}})=26\]
C) \[{{p}_{5}}=11\]
D) \[{{p}_{5}}={{p}_{2}}\cdot {{p}_{3}}\]
Correct Answer: D
Solution :
| [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 \[{{p}_{2}}={{p}_{0}}+{{p}_{1}}=2+1=3\] |
| \[{{p}_{3}}={{p}_{2}}+{{p}_{1}}=3+1=4\] |
| \[{{p}_{4}}={{p}_{3}}+{{p}_{2}}=4+3=7\] |
| \[{{p}_{5}}={{p}_{4}}+{{p}_{3}}=7+4=11\] |
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