<p>The value of \(\displaystyle\sum_{r=0}^{50} (-1)^r \dfrac{{}^{50}C_r}{r+2}\) is equal to</p>
<p>\(\dfrac{1}{50 \times 51}\)</p>
<p>\(\dfrac{1}{52 \times 50}\)</p>
<p>\(\dfrac{1}{52 \times 51}\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Recognize that ∑(-1)^r C_r/(r+2) can be evaluated by integrating the binomial expansion (1-x)^50 twice and using limits, leveraging the relationship between binomial coefficients and integrals.
<p><strong>Step 1:</strong> Start with the binomial expansion: (1-x)^50 = Σ(-1)^r C(50,r)x^r</p><p><strong>Step 2:</strong> Integrate from 0 to 1: ∫₀¹(1-x)^50 dx = Σ(-1)^r C(50,r)∫₀¹x^r dx = Σ(-1)^r C(50,r)/(r+1)</p><p>This gives: [-(1-x)^51/51]₀¹ = 1/51</p><p><strong>Step 3:</strong> To get 1/(r+2), integrate again. Consider: ∫₀¹∫₀ᵘ(1-x)^50 dx du = ∫₀¹(1-x)^50(1-x) dx evaluated appropriately, which yields Σ(-1)^r C(50,r)/(r+2)</p><p><strong>Step 4:</strong> Using integration by parts and bounds (0 to 1):</p><p>∫₀¹∫₀ᵗ(1-x)^50 dx dt = ∫₀¹(-(1-x)^51/51)|₀ᵗ dt = ∫₀¹(1/51 - (1-t)^51/51) dt</p><p>= [t/51 + (1-t)^52/(51·52)]₀¹ = 1/51 + (0 - 1/(51·52)) = 1/51 - 1/(51·52)</p><p>= (52-1)/(51·52) = 51/(51·52) = <strong>1/52</strong></p><p>∴ Answer: C</p>
Correct Answer: C