Sequences & Series
Sum to infinity
Grade 11
Question:
<p>If <em>S</em> denotes the sum to infinity and \(S_n\) the sum of <em>n</em> terms of the series \(1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \cdots\), such that \(S - S_n < \dfrac{1}{1000}\), then the least value of <em>n</em> is</p>
<p>(1) 8</p>
<p>(2) 9</p>
<p>(3) 10</p>
<p>(4) 11</p>
Step-by-Step Solution
Key Concept: Recognize this as a geometric series with first term a=1 and common ratio r=1/2, then use the formula S = a/(1-r) and S_n = a(1-r^n)/(1-r) to find the remainder S - S_n = ar^n/(1-r).
<p><strong>Step 1:</strong> Identify the series as geometric with a = 1 and r = 1/2 (since each term is half the previous).</p><p><strong>Step 2:</strong> Sum to infinity: S = a/(1-r) = 1/(1-1/2) = 1/(1/2) = 2</p><p><strong>Step 3:</strong> Sum of n terms: S_n = a(1-r^n)/(1-r) = 1·(1-(1/2)^n)/(1/2) = 2(1-(1/2)^n) = 2 - 2·(1/2)^n</p><p><strong>Step 4:</strong> Find S - S_n = 2 - [2 - 2·(1/2)^n] = 2·(1/2)^n = (1/2)^(n-1)</p><p><strong>Step 5:</strong> Simplify: S - S_n = 1/2^(n-1) or equivalently 2/2^n = (1/2)^(n-1)</p><p>∴ Answer: C</p>
Correct Answer: C