Sequences & Series
Recursive Functions and Summation
Grade 11
Question:
<p>Let <i>f</i> be defined on the natural numbers as follows: <i>f</i>(1) = 1 and for <i>n</i> ≥ 1, <i>f</i>(<i>n</i>) = <i>f</i>[<i>f</i>(<i>n</i> − 1)] + <i>f</i>[<i>n</i> − <i>f</i>(<i>n</i> − 1)]. The value of \(\sum_{r=1}^{30} f(r)\) is</p>
Step-by-Step Solution
Key Concept: Recognize the recursive definition and compute initial terms to identify that f(n) = n, then use the formula for sum of first n natural numbers.
<p><strong>Step 1:</strong> We have <i>f</i>(1) = 1.</p><p><strong>Step 2:</strong> For <i>n</i> = 2: <i>f</i>(2) = <i>f</i>[<i>f</i>(1)] + <i>f</i>[2 − <i>f</i>(1)] = <i>f</i>(1) + <i>f</i>(1) = 1 + 1 = 2.</p><p><strong>Step 3:</strong> For <i>n</i> = 3: <i>f</i>(3) = <i>f</i>[<i>f</i>(2)] + <i>f</i>[3 − <i>f</i>(2)] = <i>f</i>(2) + <i>f</i>(1) = 2 + 1 = 3.</p><p><strong>Step 4:</strong> Computing further terms reveals the pattern: <i>f</i>(<i>n</i>) = <i>n</i> for all <i>n</i>.</p><p><strong>Step 5:</strong> Therefore, $\sum_{r=1}^{30} f(r) = \sum_{r=1}^{30} r = \frac{30 \times 31}{2} = 465$.</p>
Correct Answer: 7