<p>\(\displaystyle\sum_{k=1}^{\infty} k\left(1-\dfrac{1}{n}\right)^{k-1} =\)</p>
Step-by-Step Solution
Key Concept: Recognize this as the derivative of a geometric series. If S = Σx^k, then S' = Σkx^(k-1). Use x = (1-1/n) and apply the geometric series formula for differentiation.
<p><strong>Step 1:</strong> Recognize the sum structure. Let x = (1 - 1/n), so we need: ∑_{k=1}^{∞} k·x^{k-1}</p><p><strong>Step 2:</strong> Recall that for geometric series: ∑_{k=0}^{∞} x^k = 1/(1-x) for |x| < 1</p><p><strong>Step 3:</strong> Differentiate both sides with respect to x: d/dx[∑_{k=0}^{∞} x^k] = d/dx[1/(1-x)]</p><p>This gives: ∑_{k=1}^{∞} k·x^{k-1} = 1/(1-x)²</p><p><strong>Step 4:</strong> Substitute x = (1 - 1/n) = (n-1)/n:</p><p>∑_{k=1}^{∞} k(1-1/n)^{k-1} = 1/[1-(n-1)/n]² = 1/(1/n)² = n²</p><p><strong>Note:</strong> This converges when |1 - 1/n| < 1, which holds for n > 1.</p><p>∴ Answer: <strong>n²</strong></p>
Correct Answer: C