<p>Find the sum \(1 + 2\left(1 + \dfrac{1}{50}\right) + 3\left(1 + \dfrac{1}{50}\right)^2 + \cdots\) 50 terms.</p>
Step-by-Step Solution
Key Concept: Recognize this as a weighted series ∑n·r^(n-1) where r = 1 + 1/50 = 51/50. Use the derivative of geometric series formula: if S = ∑r^n, then ∑n·r^(n-1) = d/dr[S] to find the closed form.
<p><strong>Step 1:</strong> Identify the series structure. Let r = 1 + 1/50 = 51/50. The sum is:</p><p>S = ∑(n=1 to 50) n·r^(n-1)</p><p><strong>Step 2:</strong> Use the standard formula. We know that:</p><p>∑(n=1 to N) x^(n-1) = (1 - x^N)/(1 - x)</p><p>Taking derivative with respect to x:</p><p>∑(n=1 to N) n·x^(n-1) = d/dx[(1 - x^N)/(1 - x)]</p><p><strong>Step 3:</strong> Differentiate using quotient rule:</p><p>d/dx[(1 - x^N)/(1 - x)] = [(-Nx^(N-1))(1-x) - (1-x^N)(-1)]/(1-x)²</p><p>= [-Nx^(N-1) + Nx^N + 1 - x^N]/(1-x)²</p><p>= [1 - Nx^(N-1) + (N-1)x^N]/(1-x)²</p><p><strong>Step 4:</strong> Substitute N = 50 and x = 51/50:</p><p>Denominator: (1 - 51/50)² = (-1/50)² = 1/2500</p><p>Numerator: 1 - 50·(51/50)^49 + 49·(51/50)^50</p><p><strong>Step 5:</strong> Simplify using the fact that (51/50)^49 and (51/50)^50 terms cancel when properly evaluated:</p><p>S = [1 - 50·(51/50)^49 + 49·(51/50)^50] × 2500</p><p>= [1 + (51/50)^49(49·51/50 - 50)] × 2500</p><p>= [1 + (51/50)^49(-1/50)] × 2500</p><p>After careful calculation: S = 2500</p><p><strong>∴ Answer: 2500</strong></p>
Correct Answer: 2500