Let ⟨ aₙ⟩ be a sequence such that a₀=0, a₁=1/2 and 2 aₙ₊₂=5 aₙ₊₁-3 aₙ, n=0,1,2,3, …. Then Σₖ₌₁¹⁰⁰ aₖ is equal to

Solution

Maths sir · 4 steps

Solve the recurrence first: aₙ = (3/2)ⁿ - 1, then the sum is just a GP minus 100.

1. Characteristic equation. Try aₙ = xⁿ in 2aₙ₊₂ = 5aₙ₊₁ - 3aₙ. This gives 2x² - 5x + 3 = 0, so (x-1)(2x-3) = 0 and x = 1 or x = 3/2.
2. Fix the constants. So aₙ = A + B(3/2)ⁿ. From a₀ = 0 we get A + B = 0. From a₁ = 1/2 we get A + 3/2B = 1/2, which gives B = 1 and A = -1. Check: a₂ = 9/4 - 1 = 5/4, which matches the recurrence.

   Σₖ₌₁¹⁰⁰ aₖ = Σₖ₌₁¹⁰⁰(3/2)ᵏ - 100 = (3/2((3/2)¹⁰⁰ - 1))/(3/2 - 1) - 100

3. Simplify. The fraction is 3((3/2)¹⁰⁰ - 1), and the bracket is exactly a₁₀₀. So the sum is 3a₁₀₀ - 100.
4. So. Σₖ₌₁¹⁰⁰ aₖ = 3a₁₀₀ - 100, which is option (b).

The idea to keep. For a linear recurrence like p aₙ₊₂ + q aₙ₊₁ + r aₙ = 0, solve the quadratic, write aₙ in closed form, and fix the constants from the first two terms. Then sum each part separately.

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