Let Cᵣ denote the binomial coefficient of xʳ in the expansion of (1 + x)¹⁰. If for α, β ∈ R, C₁ + 3.2C₂ + 5.3C₃ + ....... upto 10 terms = (α × 2¹¹)/(2^β - 1)( C₀ + C₁/2 + C₂/3 + ..... upto 10 terms ) then the value of α + β is equal to ___________.

Solution

Maths sir · 5 steps

Write the left side as Σ(2r-1)rCᵣ and read the bracket as ∫₀¹(1+x)¹⁰dx.

1. General term. The terms are 1· 1· C₁, 3· 2· C₂, 5· 3· C₃, …, so the r-th term is (2r-1) r Cᵣ for r = 1 to 10. Split it as 2r²Cᵣ - rCᵣ.
2. Standard sums. For n = 10: Σ rCᵣ = n 2ⁿ⁻¹ = 10 × 2⁹ = 5120 and Σ r²Cᵣ = n(n+1) 2ⁿ⁻² = 110 × 2⁸ = 28160. So the left side is 2 × 28160 - 5120 = 51200 = 25 × 2¹¹.
3. The bracket. The series C₀ + C₁/2 + C₂/3 + … is intended as the full series up to C₁₀/11. That full series is ∫₀¹ (1+x)¹⁰ dx.

   Σᵣ₌₀¹⁰Cᵣ/(r+1) = ∫₀¹ (1+x)¹⁰ dx = (2¹¹-1)/11

4. Match the form. Now the equation reads 25 × 2¹¹ = (α × 2¹¹)/(2^β - 1)·(2¹¹-1)/11. The factor 2^β - 1 is meant to cancel 2¹¹-1, so β = 11. That leaves α/11 = 25, so α = 275.
5. So. α + β = 275 + 11 = 286.

The common slip: 275. Reporting α alone gives 275. The question asks for α + β, so add β = 11.

The idea to keep. A Cᵣ divided by (r+1) means integrate (1+x)ⁿ. A Cᵣ times r or r² means use n2ⁿ⁻¹ and n(n+1)2ⁿ⁻². Spot which one, then write the sum in closed form.

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