Suppose Σᵣ₌₀²⁰²³ r² ²⁰²³Cᵣ = 2023 × α × 2²⁰²² . Then the value of α is ___________

Solution

Maths sir · 4 steps

Split r² as r(r-1) + r, and each piece collapses to a clean power of 2.

1. Break up r squared. Write r² = r(r-1) + r. Each piece pairs neatly with ⁿCᵣ, because r ⁿCᵣ = n ⁿ⁻¹Cᵣ₋₁ and r(r-1) ⁿCᵣ = n(n-1) ⁿ⁻²Cᵣ₋₂.
2. Sum each piece. Summing over r gives Σ r ⁿCᵣ = n 2ⁿ⁻¹ and Σ r(r-1) ⁿCᵣ = n(n-1) 2ⁿ⁻². Adding them, n 2ⁿ⁻²[(n-1) + 2] = n(n+1) 2ⁿ⁻².

   Σᵣ₌₀ⁿ r² ⁿCᵣ = n(n+1) 2ⁿ⁻²

3. Put n = 2023. The sum is 2023 × 2024 × 2²⁰²¹. Write 2024 = 1012 × 2, so it becomes 2023 × 1012 × 2²⁰²².
4. So. Matching with 2023 × α × 2²⁰²² gives α = 1012.

The common slip: 2024. Matching 2023 × 2024 × 2²⁰²¹ directly, without turning 2²⁰²¹ into 2²⁰²², gives 2024 instead of 1012.

The idea to keep. Whenever r² or r³ sits beside ⁿCᵣ, rewrite it using r, r(r-1), r(r-1)(r-2). Each of these falls straight into n, n(n-1), ... times a power of 2.

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