Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>Suppose that \(F(n+1) = \dfrac{2F(n)+1}{2}\) for \(n = 1, 2, 3, \ldots\) and \(F(1) = 2\). Then \(F(101)\) equals</p>
<p>50</p>
<p>52</p>
<p>54</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Simplify the recurrence relation to F(n+1) = F(n) + 1/2, then recognize this as an arithmetic sequence with common difference 1/2. Use the arithmetic sequence formula to find F(101) = F(1) + 100(1/2).
<p><strong>Step 1:</strong> Simplify the recurrence relation.</p><p>F(n+1) = (2F(n)+1)/2 = F(n) + 1/2</p><p><strong>Step 2:</strong> Recognize this as an arithmetic sequence with first term F(1) = 2 and common difference d = 1/2.</p><p><strong>Step 3:</strong> Use the arithmetic sequence formula: F(n) = F(1) + (n-1)d</p><p>F(101) = 2 + (101-1) × (1/2) = 2 + 100 × (1/2) = 2 + 50 = 52</p><p>∴ Answer: B (F(101) = 52)</p>
Correct Answer: B