What is \Sigma_{i}^{n}x_{i}f_{i} equal to ?

Consider the following for the next two (02) items that follow : <table> <thead><tr><th>x_i</th><th>1</th><th>2</th><th>3</th><th>...</th><th>n</th></tr></thead> <tbody> <tr><td>f_i</td><td>1</td><td>2^{-1}</td><td>2^{-2}</td><td>...</td><td>2^{-(n-1)}</td></tr> </tbody> </table>

  1. A. \frac{2^{n+1}-n+2}{2^{n-1}}
  2. B. \frac{2^{n+1}-n-2}{2^{n-1}} ✓
  3. C. \frac{2^{n+1}-n-2}{2^{n}-1}
  4. D. \frac{2^{n+1}-n+2}{2^{n}}

Correct Answer: B. \frac{2^{n+1}-n-2}{2^{n-1}}

Explanation

Let the sum be S = 1(1) + 2(1/2) + 3(1/2^2) + \dots + n(1/2^{n-1}). This is an Arithmetico-Geometric Progression (AGP). Multiply by \frac{1}{2}: \frac{S}{2} = 1(1/2) + 2(1/2^2) + \dots + n(1/2^n). Subtracting the two equations gives \frac{S}{2} = 1 + \frac{1}{2} + \frac{1}{2^2} + \dots + \frac{1}{2^{n-1}} - \frac{n}{2^n}. The terms up to the n-th form a Geometric Progression with sum \frac{1-(1/2)^n}{1-1/2} = 2(1 - \frac{1}{2^n}). Thus, \frac{S}{2} = 2 - \frac{2}{2^n} - \frac{n}{2^n} = \frac{2^{n+1} - 2 - n}{2^n}. Multiplying by 2 gives S = \frac{2^{n+1} - n - 2}{2^{n-1}}.

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