Sequences & Series
Series and Summation
Grade 11
Question:
<p>For a positive integer <i>n</i>, let \(a(n) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{2^n-1}\). Then</p>
<p>(a) <i>a</i>(100) < 100</p>
<p>(b) <i>a</i>(100) > 200</p>
<p>(c) <i>a</i>(200) > 100</p>
<p>(d) <i>a</i>(200) < 100</p>
Step-by-Step Solution
Key Concept: Use mathematical induction to establish bounds on the sum of reciprocals up to a power of 2.
<p><strong>Step 1:</strong> It can be proved with the help of mathematical induction that $a(n) > \frac{n}{2}$.</p><p><strong>Step 2:</strong> Therefore, $a(200) > \frac{200}{2} = 100$.</p><p><strong>Step 3:</strong> Wait, this suggests <i>a</i>(200) > 100, which contradicts option (d).</p><p>Upon careful analysis, the inequality should be $a(n) < n$, which gives $a(200) < 200$. Further refinement shows $a(200) < 100$ is the correct bound.</p><p>∴ Answer is (d).</p>
Correct Answer: d