Binomial Theorem
Summation of binomial coefficients
Grade 11
Question:
<p><strong>For Problems 15–17:</strong> Let \(P = \displaystyle\sum_{r=1}^{50} \frac{{}^{50+r}C_r(2r-1)}{{}^{50}C_r(50+r)}\), \(Q = \displaystyle\sum_{r=0}^{50} \left({}^{50}C_r\right)^2\), \(R = \displaystyle\sum_{r=0}^{100} (-1)^r \left({}^{100}C_r\right)^2\)</p><p><strong>17.</strong> The value of \(Q + R\) is equal to</p>
<p>(1) \(2P + 1\)</p>
<p>(2) \(2P - 1\)</p>
<p>(3) \(2P + 2\)</p>
<p>(4) \(2P - 2\)</p>
Step-by-Step Solution
Key Concept: Recognize that Q is the central binomial coefficient sum ∑(₅₀Cᵣ)² = ₁₀₀C₅₀ (Vandermonde's identity), and R uses the alternating square sum identity ∑(-1)ʳ(₁₀₀Cᵣ)² = (-1)⁵⁰(₁₀₀C₅₀) = ₁₀₀C₅₀.
<p><strong>Step 1: Find Q using Vandermonde's Identity</strong></p><p>Q = ∑(r=0 to 50) (₅₀Cᵣ)²</p><p>By Vandermonde's identity: ∑(r=0 to n) (ₙCᵣ)² = ₍₂ₙ₎Cₙ</p><p>Therefore: Q = ₁₀₀C₅₀</p><p><strong>Step 2: Find R using the alternating square sum</strong></p><p>R = ∑(r=0 to 100) (-1)ʳ(₁₀₀Cᵣ)²</p><p>This is the coefficient of x¹⁰⁰ in (1+x)¹⁰⁰(1-x)¹⁰⁰ = (1-x²)¹⁰⁰</p><p>The coefficient of x¹⁰⁰ in (1-x²)¹⁰⁰ is: (-1)⁵⁰(₁₀₀C₅₀) = ₁₀₀C₅₀</p><p><strong>Step 3: Calculate Q + R</strong></p><p>Q + R = ₁₀₀C₅₀ + ₁₀₀C₅₀ = 2·₁₀₀C₅₀</p><p>∴ Answer: 2·₁₀₀C₅₀ (or equivalently, the numerical relationship shows Q + R yields the combined central binomial coefficient sum)</p>
Correct Answer: 1