Let a₁=b₁=1 and aₙ = aₙ₋₁ + (n - 1),bₙ = bₙ₋₁ + aₙ₋₁,∀ n ≥ 2. If S = Σₙ₌₁¹⁰ bₙ/2ⁿ and T = Σₙ₌₁⁸ n/2ⁿ⁻¹, then 2⁷(2S - T) is equal to ____________.

Solution

Maths sir · 4 steps

Halve the sum and subtract it. Differences of b give a, and differences of a give n.

1. List the terms. aₙ = 1 + n(n-1)/2 gives a₉ = 37. Building bₙ step by step gives 1, 2, 4, 8, 15, 26, 42, 64, 93, 130, so b₁₀ = 130.
2. Shift and subtract. Write S - S/2 by pairing each bₙ/2ⁿ with bₙ₋₁/2ⁿ. The numerators become bₙ - bₙ₋₁ = aₙ₋₁, and one leftover tail term remains. This gives S = 1 + U - 130/2¹⁰, where U = Σₖ₌₁⁹ aₖ/2ᵏ.
3. Repeat on U. The same trick on U turns the numerators into aₖ - aₖ₋₁ = k-1. The sum Σₘ₌₁⁸ m/2ᵐ⁺¹ is exactly T/4. So U = 1 + T/2 - 37/2⁹.

   2S = 4 + T - 2(74 + 130)/1024 = 4 + T - 51/128

4. So. 2S - T = 4 - 51/128 = 461/128. Multiplying by 2⁷ = 128 gives 461.

The common slip: 512. Dropping the leftover last terms 37/2⁹ and 130/2¹⁰ makes 2S - T exactly 4, which gives 512 instead of 461.

The idea to keep. For a sum like Σ bₙ/2ⁿ, subtract half of it from itself. This turns bₙ into its differences. Repeat until the numerator is linear, and always carry the leftover last terms.

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