If Σᵣ₌₁⁹((r+3)/2ʳ) · ⁹ Cᵣ=α(3/2)⁹-β, α, β ∈ ℕ, then (α+β)² is equal to

Solution

Maths sir · 4 steps

Split (r+3)/2ʳ into two pieces, and each one collapses to a power of 3/2.

1. Split the sum. Write (r+3)/2ʳ = r(1/2)ʳ + 3(1/2)ʳ. The sum becomes S₁ + 3S₂, where S₁ = Σᵣ₌₁⁹ r ⁹Cᵣ(1/2)ʳ and S₂ = Σᵣ₌₁⁹ ⁹Cᵣ(1/2)ʳ.
2. First sum. Use r ⁹Cᵣ = 9 ⁸Cᵣ₋₁. Then S₁ = 9/2Σₛ₌₀⁸ ⁸Cₛ(1/2)ˢ = 9/2(3/2)⁸ = 3(3/2)⁹.
3. Second sum. The full expansion from r = 0 gives (1+1/2)⁹. Here the sum starts at r = 1, so remove the r = 0 term, which is 1: S₂ = (3/2)⁹ - 1.

   S₁ + 3S₂ = 3(3/2)⁹ + 3(3/2)⁹ - 3 = 6(3/2)⁹ - 3

4. So. Comparing the two sides gives α = 6 and β = 3, so (α+β)² = 9² = 81, option (b).

The idea to keep. When a binomial sum has r multiplying ⁿCᵣ, use r ⁿCᵣ = n ⁿ⁻¹Cᵣ₋₁ to remove it. Always check where the sum starts and subtract any missing r = 0 term.

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