Quadratic Equation Question 3
Question 3 - 2024 (27 Jan Shift 2)
If $\alpha, \beta$ are the roots of the equation, $x^{2}-x-1=0$ and $S_{n}=2023 \alpha^{n}+2024 \beta^{n}$, then
(1) $2 \mathrm{~S}{12}=\mathrm{S}{11}+\mathrm{S}_{10}$
(2) $\mathrm{S}{12}=\mathrm{S}{11}+\mathrm{S}_{10}$
(3) $2 \mathrm{~S}{11}=\mathrm{S}{12}+\mathrm{S}_{10}$
(4) $S_{11}=S_{10}+S_{12}$
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Answer (2)
Solution
$\mathrm{x}^{2}-\mathrm{x}-1=0$
$\mathrm{S}_{\mathrm{n}}=2023 \alpha^{\mathrm{n}}+2024 \beta^{\mathrm{n}}$
$\mathrm{S}{\mathrm{n}-1}+\mathrm{S}{\mathrm{n}-2}=2023 \alpha^{\mathrm{n}-1}+2024 \beta^{\mathrm{n}-1}+2023 \alpha^{\mathrm{n}-2}+2024 \beta^{\mathrm{n}-2}$
$=2023 \alpha^{\mathrm{n}-2}[1+\alpha]+2024 \beta^{\mathrm{n}-2}[1+\beta]$
$=2023 \alpha^{\mathrm{n}-2}\left[\alpha^{2}\right]+2024 \beta^{\mathrm{n}-2}\left[\beta^{2}\right]$
$=2023 \alpha^{\mathrm{n}}+2024 \beta^{\mathrm{n}}$
$\mathrm{S}{\mathrm{n}-1}+\mathrm{S}{\mathrm{n}-2}=\mathrm{S}_{\mathrm{n}}$
Put $\mathrm{n}=12$
$\mathrm{S}{11}+\mathrm{S}{10}=\mathrm{S}_{12}$
