<p>If \(\displaystyle\sum_{r=0}^{n}(pr+2) \cdot {}^nC_r = (25)(64)\) where \(n, p \in N\), then</p>
Step-by-Step Solution
Key Concept: Split the sum into two parts using linearity: p∑r·ⁿCᵣ + 2∑ⁿCᵣ. The first sum equals n·2ⁿ⁻¹ (using r·ⁿCᵣ = n·ⁿ⁻¹Cᵣ₋₁) and the second equals 2ⁿ, giving p·n·2ⁿ⁻¹ + 2·2ⁿ = 25·64.
<p><strong>Step 1:</strong> Decompose the sum using linearity:</p><p>∑(pr + 2)·ⁿCᵣ = p∑r·ⁿCᵣ + 2∑ⁿCᵣ</p><p><strong>Step 2:</strong> Evaluate standard binomial sums:</p><p>• ∑ⁿCᵣ = 2ⁿ (standard result)</p><p>• ∑r·ⁿCᵣ = n·2ⁿ⁻¹ (using r·ⁿCᵣ = n·ⁿ⁻¹Cᵣ₋₁)</p><p><strong>Step 3:</strong> Substitute into the equation:</p><p>p·n·2ⁿ⁻¹ + 2·2ⁿ = 25·64</p><p>p·n·2ⁿ⁻¹ + 2ⁿ⁺¹ = 1600</p><p><strong>Step 4:</strong> Factor and simplify:</p><p>2ⁿ⁻¹(pn + 4) = 1600 = 25 × 64 = 25 × 2⁶</p><p>2ⁿ⁻¹(pn + 4) = 25 × 2⁶</p><p><strong>Step 5:</strong> Match powers of 2:</p><p>Since 25 is odd, we need: n - 1 = 6, so <strong>n = 7</strong></p><p>Then: pn + 4 = 25, so 7p + 4 = 25, giving <strong>p = 3</strong></p><p>∴ Answer: n = 7, p = 3 (Options A, D)
Correct Answer: A,D