Sequences & Series
Sum of Series
Grade 11
Question:
<p>313. Sum of the first \(n\) terms of the series \(\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \ldots\) is equal to</p>
<p>(a) \(2^n - n - 1\)</p>
<p>(b) \(1 - 2^{-n}\)</p>
<p>(c) \(n + 2^{-n} - 1\)</p>
<p>(d) \(2^n - 1\)</p>
Step-by-Step Solution
Key Concept: Recognize that each term has the form (2^k - 1)/2^k = 1 - 1/2^k. The sum telescopes when you separate it into a linear part and a geometric series of reciprocals.
<p><strong>Step 1: Identify the general term</strong></p><p>The series is: 1/2 + 3/4 + 7/8 + 15/16 + ...</p><p>Notice: 1/2 = 1 - 1/2, 3/4 = 1 - 1/4, 7/8 = 1 - 1/8, 15/16 = 1 - 1/16</p><p>General term: T_k = (2^k - 1)/2^k = 1 - 1/2^k</p><p><strong>Step 2: Find the sum of first n terms</strong></p><p>S_n = Σ(k=1 to n) (1 - 1/2^k)</p><p>S_n = Σ(k=1 to n) 1 - Σ(k=1 to n) 1/2^k</p><p>S_n = n - [1/2 + 1/4 + 1/8 + ... + 1/2^n]</p><p><strong>Step 3: Evaluate the geometric series</strong></p><p>The geometric series with first term a = 1/2, ratio r = 1/2:</p><p>Σ(k=1 to n) 1/2^k = (1/2)(1 - (1/2)^n)/(1 - 1/2) = 1 - 1/2^n</p><p><strong>Step 4: Final answer</strong></p><p>S_n = n - (1 - 1/2^n) = n - 1 + 1/2^n</p><p>∴ Answer: n - 1 + 1/2^n (or equivalently: n - (2^n - 1)/2^n)</p>
Correct Answer: C