Let ⁿCᵣ denote the binomial coefficient of xʳ in the expansion of (1 + x)ⁿ. If Σₖ₌₀¹⁰ (2² + 3k) ¹⁰Cₖ = α .3¹⁰ + β .2¹⁰,α ,β ∈ R, then α + β is equal to ___________.

Solution

Maths sir · 4 steps

Split the sum into a constant part and a k part. The 3¹⁰ term never shows up, so α = 0.

1. Split the sum. The bracket is 4 + 3k, so the sum breaks into 4Σ ¹⁰Cₖ + 3Σ k ¹⁰Cₖ, with k running from 0 to 10.
2. First part. Putting x = 1 in (1+x)¹⁰ gives Σ ¹⁰Cₖ = 2¹⁰. So the first part is 4 × 2¹⁰.
3. Second part. Use k ⁿCₖ = n ⁿ⁻¹Cₖ₋₁. Then Σ k ¹⁰Cₖ = 10Σ ⁹Cₖ₋₁ = 10 × 2⁹ = 5 × 2¹⁰, so the second part is 3 × 5 × 2¹⁰ = 15 × 2¹⁰.

   Σₖ₌₀¹⁰(4+3k) ¹⁰Cₖ = 4 × 2¹⁰ + 15 × 2¹⁰ = 19 × 2¹⁰

4. So. Matching with α · 3¹⁰ + β · 2¹⁰ gives α = 0 and β = 19, so α + β = 19.

The common slip: 16. Reading 2² as 2ᵏ gives 3¹⁰ + 15 × 2¹⁰, so α = 1, β = 15 and the sum 16. The term is the constant 4, so no 3¹⁰ appears.

The idea to keep. For sums like Σ (a + bk) ⁿCₖ, use Σ ⁿCₖ = 2ⁿ and Σ k ⁿCₖ = n 2ⁿ⁻¹. A 3ⁿ term only appears if 2ᵏ sits inside the sum.

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