Sankalp JEE Full Test-1 Question-10 Solution

Question 10. 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 S_{12}=S_{11}+S_{10}

(2) S_{12}=S_{11}+S_{10}

(3) 2 S_{11}=S_{12}+S_{10}

(4) S_{11}=S_{10}+S_{12}

Answer (2)

Explanation:

x^{2}-x-1=0 S_{n}=2023 \alpha^{n}+2024 \beta^{n} S_{n-1}+S_{n-2}=2023 \alpha^{n-1}+2024 \beta^{n-1}+2023 \alpha^{n-2}+2024 \beta^{n-2} =2023 \alpha^{n-2}[1+\alpha]+2024 \beta^{n-2}[1+\beta] =2023 \alpha^{n-2}\left[\alpha^{2}\right]+2024 \beta^{n-2}\left[\beta^{2}\right] =2023 \alpha^{n}+2024 \beta^{n} S_{n-1}+S_{n-2}=S_{n}

Put n=12

S_{11}+S_{10}=S_{12}
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