Let { aₙ }ₙ₌₁^∞ be a sequence such that a₁ = 1, a₂ = 1 and aₙ₊₂ = 2aₙ₊₁ + aₙ for all n ≥ 1. Then the value of 47Σₙ₌₁^∞ aₙ/2³ⁿ is equal to ______________.

Solution

Maths sir · 4 steps

Call the sum S, then use the recurrence to wipe out almost every term of S - 2xS - x²S.

1. Name the sum. Since 2³ⁿ = 8ⁿ, the sum is S = Σ aₙ xⁿ with x = 1/8. The roots of t² = 2t + 1 are 1 ± √2, and 1 + √2 < 8. So the terms shrink fast enough for the series to converge.
2. Shift and subtract. Form S - 2xS - x²S. For every n ≥ 3 the coefficient of xⁿ is aₙ - 2aₙ₋₁ - aₙ₋₂ = 0 by the recurrence. Only the first two powers survive: x¹ has coefficient a₁ = 1, and x² has coefficient a₂ - 2a₁ = -1.

   S(1 - 2x - x²) = x - x²

3. Put in numbers. With x = 1/8, the right side is 1/8 - 1/64 = 7/64. The bracket is 1 - 16/64 - 1/64 = 47/64. So S = 7/47.
4. So. 47S = 47 × 7/47 = 7, so the answer is 7.

The common slip: 9. Missing the -2a₁ in the x² coefficient makes the right side x + x² = 9/64, and then 47S comes out as 9.

The idea to keep. For Σ aₙ xⁿ with a linear recurrence, multiply S by the recurrence's coefficients and subtract. Every term cancels except the first few, so read those off carefully from a₁ and a₂.

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