Let α, β ; α>β, be the roots of the equation x²-√2 x-√3=0. Let Pₙ=αⁿ-βⁿ, n ∈ N. Then (11 √3-10 √2) P₁₀+(11 √2+10) P₁₁-11 P₁₂ is equal to

Solution

Maths sir · 4 steps

Both roots satisfy x² = √2x + √3, so Pₙ obeys the same recurrence. Use it twice.

1. Build the recurrence. Since α² = √2α + √3, multiply by αⁿ to get αⁿ⁺² = √2αⁿ⁺¹ + √3αⁿ. The same holds for β. Subtract the two equations.

   Pₙ₊₂ = √2 Pₙ₊₁ + √3 Pₙ

2. Remove P12. Put P₁₂ = √2P₁₁ + √3P₁₀. Then -11P₁₂ = -11√2P₁₁ - 11√3P₁₀, which cancels the 11√2P₁₁ and 11√3P₁₀ terms.
3. What is left. The expression becomes 10P₁₁ - 10√2P₁₀ = 10(P₁₁ - √2P₁₀). By the recurrence with n = 9, P₁₁ - √2P₁₀ = √3P₉.
4. So. The expression equals 10√3 P₉, which is option (a).

The idea to keep. When Pₙ = αⁿ - βⁿ and α, β are roots of x² = px + q, then Pₙ₊₂ = pPₙ₊₁ + qPₙ. Use this to reduce higher terms step by step.

Answer from the NTA final answer key. Solution by PrepWiser.