Let { aₙ} ₙ₌₀^∞ be a sequence such that a₀ = a₁ = 0 and aₙ₊₂ = 2aₙ₊₁ - aₙ + 1 for all n ≥ 0. Then, Σₙ₌₂^∞ aₙ/7ⁿ is equal to:

Solution

Maths sir · 4 steps

Do not hunt for every term: multiply the recurrence by powers of 1/7, add, and S solves itself.

1. Name the sum. Let x = 1/7 and S = Σₙ₌₂^∞ aₙ xⁿ. Since a₀ = a₁ = 0, adding the n = 0 and n = 1 terms changes nothing, so S = Σₙ₌₀^∞ aₙ xⁿ.
2. Use the recurrence. Multiply aₙ₊₂ = 2aₙ₊₁ - aₙ + 1 by xⁿ⁺² and add over all n ≥ 0. The left side gives S - a₀ - a₁ x = S. On the right, the three sums are 2xS, x² S and x²(1 + x + x² + …) = x²/(1-x).

   S = 2xS - x² S + x²/(1-x) ⇒ S(1-x)² = x²/(1-x) ⇒ S = x²/(1-x)³

3. Put in x. With x = 1/7: x² = 1/49 and (1-x)³ = (6/7)³ = 216/343. So S = 1/49 × 343/216 = 7/216.
4. So. The sum is 7/216, which is option (b). As a check, the terms are aₙ = n(n-1)/2, that is 0, 0, 1, 3, 6, …, and Σ C(n, 2)xⁿ = x²/(1-x)³ gives the same value.

The idea to keep. When a sequence is defined by a linear recurrence and you need Σ aₙ xⁿ, multiply the recurrence by xⁿ and add. Every sum turns back into S, and you solve one linear equation.

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