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Quadratic Equation and Inequalities question

2023 · 11 Apr · Shift 1 · Q47
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Quadratic Equation and Inequalities question

2023 · 11 Apr · Shift 1 · Q47

JEE MainMathematicsQuadratic Equation and InequalitiesNumerical+4 / −1
If aaa and bbb are the roots of the equation x2−7x−1=0x^{2}-7 x-1=0x2−7x−1=0, then the value of a21+b21+a17+b17a19+b19\frac{a^{21}+b^{21}+a^{17}+b^{17}}{a^{19}+b^{19}}a19+b19a21+b21+a17+b17​ is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 51

  1. Let Sn=an+bn.S_n=a^n+b^n.Sn​=an+bn. We need to find a21+b21+a17+b17a19+b19=S21+S17S19.\frac{a^{21}+b^{21}+a^{17}+b^{17}}{a^{19}+b^{19}}=\frac{S_{21}+S_{17}}{S_{19}}.a19+b19a21+b21+a17+b17​=S19​S21​+S17​​.

  2. Since a,ba,ba,b are roots of x2−7x−1=0,x^2-7x-1=0,x2−7x−1=0, each root satisfies r2=7r+1.r^2=7r+1.r2=7r+1. Multiplying by rn−2r^{n-2}rn−2, we get for n≥2n\ge 2n≥2: rn=7rn−1+rn−2.r^n=7r^{n-1}+r^{n-2}.rn=7rn−1+rn−2. This holds for both r=ar=ar=a and r=br=br=b. Adding for aaa and bbb: Sn=7Sn−1+Sn−2.S_n=7S_{n-1}+S_{n-2}.Sn​=7Sn−1​+Sn−2​.

  3. Now apply this recurrence to S21S_{21}S21​: S21=7S20+S19.S_{21}=7S_{20}+S_{19}.S21​=7S20​+S19​. Also, S20=7S19+S18.S_{20}=7S_{19}+S_{18}.S20​=7S19​+S18​. So, S21=7(7S19+S18)+S19=50S19+7S18.S_{21}=7(7S_{19}+S_{18})+S_{19}=50S_{19}+7S_{18}.S21​=7(7S19​+S18​)+S19​=50S19​+7S18​. Hence S21+S17=50S19+7S18+S17.S_{21}+S_{17}=50S_{19}+7S_{18}+S_{17}.S21​+S17​=50S19​+7S18​+S17​.

  4. Use the recurrence again for S19S_{19}S19​: S19=7S18+S17.S_{19}=7S_{18}+S_{17}.S19​=7S18​+S17​. Therefore, 7S18+S17=S19.7S_{18}+S_{17}=S_{19}.7S18​+S17​=S19​. Substituting, S21+S17=50S19+S19=51S19.S_{21}+S_{17}=50S_{19}+S_{19}=51S_{19}.S21​+S17​=50S19​+S19​=51S19​.

  5. Thus, S21+S17S19=51S19S19=51.\frac{S_{21}+S_{17}}{S_{19}}=\frac{51S_{19}}{S_{19}}=51.S19​S21​+S17​​=S19​51S19​​=51.

So the required value is 51.\boxed{51}.51​.

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