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જો $\alpha$ અને $\beta$ એ સમીકરણ $\mathrm{x}^{2}-\mathrm{x}-1=0 $ ના બીજ હોય અને $\mathrm{p}_{\mathrm{k}}=(\alpha)^{\mathrm{k}}+(\beta)^{\mathrm{k}}, \mathrm{k} \geq 1,$ તો આપેલ પૈકી ક્યૂ વિધાન સત્ય છે ?
$\left(p_{1}+p_{2}+p_{3}+p_{4}+p_{5}\right)=26$
$\mathrm{p}_{5}=11$
$\mathrm{p}_{3}=\mathrm{p}_{5}-\mathrm{p}_{4}$
$\mathrm{p}_{5}=\mathrm{p}_{2} \cdot \mathrm{p}_{3}$
Solution
$\alpha+\beta=1, \alpha \beta=-1$
$\mathrm{P}_{\mathrm{k}}=\alpha^{\mathrm{k}}+\beta^{\mathrm{k}}$
$\alpha^{2}-\alpha-1=0$
$\Rightarrow \alpha^{\mathrm{k}}-\alpha^{\mathrm{k}-1}-\alpha^{\mathrm{k}-2}=0$
and $\beta^{\mathrm{k}}-\beta^{\mathrm{k}-1}-\beta^{\mathrm{k}-2}=0$
$\Rightarrow \mathrm{P}_{\mathrm{k}}=\mathrm{P}_{\mathrm{k}-1}+\mathrm{P}_{\mathrm{k}-2}$
$P_{1}=\alpha+\beta=1$
$\mathrm{P}_{2}=(\alpha+\beta)^{2}-2 \alpha \beta=1+2=3$
$\mathrm{P}_{3}=4$
$\mathrm{P}_{4}=7$
$\mathrm{P}_{5}=11$
