Let α and β be two real numbers such that α + β = 1 and αβ = -1. Let pₙ = (α)ⁿ + (β)ⁿ, pₙ₋₁ = 11 and pₙ₊₁ = 29 for some integer n ≥ 1. Then, the value of pₙ² is ___________.

Solution

Maths sir · 4 steps

Alpha and beta are roots of a quadratic, so each power-sum is just the sum of the previous two.

1. Build the quadratic. Sum of roots is 1 and product is -1, so α and β are the roots of x² - x - 1 = 0. That gives α² = α + 1 and β² = β + 1.
2. Raise the powers. Multiply α² = α + 1 by αⁿ⁻¹ to get αⁿ⁺¹ = αⁿ + αⁿ⁻¹. Do the same for β and add the two equations.

   pₙ₊₁ = pₙ + pₙ₋₁

3. Put in numbers. 29 = pₙ + 11, so pₙ = 18. The sequence 11, 18, 29 adds up correctly.
4. So. pₙ² = 18² = 324.

The common slip: 18. Stopping at pₙ = 18 answers the wrong thing. The question asks for pₙ², which is 324.

The idea to keep. When α, β are roots of x² = ax + b, the sums pₙ = αⁿ + βⁿ follow pₙ₊₁ = a pₙ + b pₙ₋₁. Read a and b straight off the quadratic.

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