<p>Find the sum of the series
\[1+2(1-x)+3(1-x)(1-2x)+\cdots+n(1-x)(1-2x)\cdots(1-(n-1)x).\]</p>
Step-by-Step Solution
Key Concept: Recognize that each term can be written as a difference of products using telescoping series technique. Specifically, if we denote P_k = (1-x)(1-2x)⋯(1-kx), then the k-th term equals k·P_{k-1}, which can be expressed as a difference P_{k-1} - P_k when properly manipulated.
<p><strong>Step 1: Denote the product terms</strong></p><p>Let P_k = (1-x)(1-2x)(1-3x)⋯(1-kx) where P_0 = 1.</p><p>The given series is: S = 1·P_0 + 2·P_1 + 3·P_2 + ⋯ + n·P_{n-1}</p><p><strong>Step 2: Express each term as a difference</strong></p><p>We need to show that k·P_{k-1} can be written as a telescoping difference. Multiply both sides of the relation P_k = (1-kx)·P_{k-1} by differentiating concept:</p><p>P_{k-1} - P_k = P_{k-1} - (1-kx)·P_{k-1} = P_{k-1}[1-(1-kx)] = kx·P_{k-1}</p><p>Therefore: k·P_{k-1} = (1/x)[P_{k-1} - P_k]</p><p><strong>Step 3: Rewrite the series</strong></p><p>S = Σ(k=1 to n) k·P_{k-1} = Σ(k=1 to n) (1/x)[P_{k-1} - P_k]</p><p>S = (1/x)·Σ(k=1 to n) [P_{k-1} - P_k]</p><p><strong>Step 4: Apply telescoping</strong></p><p>S = (1/x)[(P_0 - P_1) + (P_1 - P_2) + (P_2 - P_3) + ⋯ + (P_{n-1} - P_n)]</p><p>Most terms cancel, leaving:</p><p>S = (1/x)[P_0 - P_n]</p><p><strong>Step 5: Substitute back the values</strong></p><p>Since P_0 = 1 and P_n = (1-x)(1-2x)⋯(1-nx):</p><p>∴ Answer: S = (1/x)[1 - (1-x)(1-2x)⋯(1-nx)] = $\dfrac{1}{x}\left[1-(1-x)(1-2x)\cdots(1-nx)\right]$</p>
Correct Answer: \(\dfrac{1}{x}\left[1-(1-x)(1-2x)\cdots(1-nx)\right]\)