If alpha, beta are the roots of the equation, | vedclass

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Question

${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}_{\ma

Vedclass · Accepted Answer

$\mathrm{S}_{12}=\mathrm{S}_{11}+\mathrm{S}_{10}$

Vedclass · Answer

${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}} $

$ \mathrm{Put} \mathrm{n}=12 $

$\mathrm{~S}_{11}+\mathrm{S}_{10}=\mathrm{S}_{12}$

If $\alpha, \beta$ are the roots of the equation, $x^2-x-1=0$ and $S_n=2023 \alpha^n+2024 \beta^n$, then

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