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Theory of Equations: Roots Equation Value Equal

JEE Maths question with a full step-by-step solution.

Question
If α,β,γ\alpha, \beta, \gamma are roots of the equation x3−2x2−1=0x^3 - 2x^2 - 1 = 0 and Tn=αn+βn+γnT_n = \alpha^n + \beta^n + \gamma^n, then the value of T11−T8T10\dfrac{T_{11} - T_8}{T_{10}} is equal to
A11
B22correct
C−1-1
D33
Solution
Step 1: From x3−2x2−1=0x^3 - 2x^2 - 1 = 0: each root satisfies α3=2α2+1\alpha^3 = 2\alpha^2 + 1, so α3−1=2α2\alpha^3 - 1 = 2\alpha^2. Step 2:
T11−T8=∑α11−∑α8=∑α8(α3−1)=∑α8⋅2α2=2∑α10=2T10T_{11} - T_8 = \sum \alpha^{11} - \sum \alpha^8 = \sum \alpha^8(\alpha^3 - 1) = \sum \alpha^8 \cdot 2\alpha^2 = 2\sum \alpha^{10} = 2T_{10}
 Step 3:
T11−T8T10=2T10T10=2\frac{T_{11} - T_8}{T_{10}} = \frac{2T_{10}}{T_{10}} = 2
 Correct answer: (2) Note: This question can also solved by newton's formula
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