Verified Solution • 2025–26 Session moderate

QJEEM24Maa42f: JEE Main 2024 Mathematics - Class 11 JEEM24Maa42f Solution: If are the roots of the equation, and , then :

If α,β\alpha, \beta are the roots of the equation, x2−x−1=0x^2-x-1=0 and Sn=2023αn+2024βnS_n=2023 \alpha^n+2024 \beta^n , then :

Multiple Choice Options

  • A 2S12=S11+S102 S_{12}=S_{11}+S_{10}
  • B S12=S11+S10S_{12}=S_{11}+S_{10}
  • C S11=S10+S12S_{11}=S_{10}+S_{12}
  • D 2S11=S12+S102 S_{11}=S_{12}+S_{10}

Step-by-Step Verified Solution

amp;x2−x−1=0amp;Sn=2023αn+2024βnamp;Sn−1+Sn−2=2023αn−1+2024βn−1+2023αn−2+2024βn−2amp;=2023αn−2[1+α]+2024βn−2[1+β]amp;=2023αn−2[α2]+2024βn−2[β2]amp;=2023αn+2024βnamp;Sn−1+Sn−2=Snamp;Putn=12amp;S11+S10=S12\begin{aligned}& 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 \\& P_{u t} n=12 \\& S_{11}+S_{10}=S_{12}\end{aligned}

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