<p>Let \(f(x) = 9^x/(9^x + 3)\) for all \(x \in \mathbb{R}\). Then the value of \(\displaystyle\sum_{r=1}^{2008} f(r/2009)\) is ______.</p>
Step-by-Step Solution
Key Concept: Recognize that f(x) + f(1-x) = 1 (a complementary property), which allows pairing terms in the sum symmetrically to collapse the entire series.
<p><strong>Step 1:</strong> Verify the complementary property. For f(x) = 9^x/(9^x + 3):</p><p>f(1-x) = 9^(1-x)/(9^(1-x) + 3) = (9/9^x)/(9/9^x + 3) = 9/(9 + 3·9^x) = 9/(3(3 + 3^(2x))) = 3/(3 + 9^x)</p><p>Therefore: f(x) + f(1-x) = 9^x/(9^x + 3) + 3/(9^x + 3) = (9^x + 3)/(9^x + 3) = 1</p><p><strong>Step 2:</strong> Apply pairing to the sum. For S = Σ(r=1 to 2008) f(r/2009):</p><p>Pair term r with term (2009-r):</p><p>f(r/2009) + f((2009-r)/2009) = f(r/2009) + f(1 - r/2009) = 1</p><p><strong>Step 3:</strong> Count the pairs. The terms r = 1,2,...,2008 pair as:</p><p>(1,2008), (2,2007), (3,2006), ..., (1004,1005)</p><p>This gives exactly 1004 pairs, each summing to 1.</p><p><strong>Step 4:</strong> Calculate total sum:</p><p>S = 1004 × 1 = 1004</p><p>∴ Answer: <strong>1004</strong></p>
Correct Answer: 1004