<p>Let <em>f</em> be a real valued function satisfying <br> \( f(x) + f(1-x) = \dfrac{9^x}{9^x+3} + \dfrac{9^{1-x}}{9^{1-x}+3} \). <br> Find the value of \( \displaystyle\sum_{r=1}^{2008} f\!\left(\frac{r}{2009}\right) \).</p>
Step-by-Step Solution
Key Concept: Use the functional equation f(x) + f(1-x) = constant by substituting x and (1-x) to find that the RHS equals 1 for all x. Then pair up terms in the sum symmetrically: f(r/2009) + f((2009-r)/2009) = 1.
<p><strong>Step 1:</strong> Simplify the RHS using the functional equation property.</p><p>Let RHS = 9^x/(9^x+3) + 9^(1-x)/(9^(1-x)+3)</p><p>Note that 9^(1-x) = 9/9^x, so:</p><p>9^(1-x)/(9^(1-x)+3) = (9/9^x)/(9/9^x + 3) = 9/(9+3·9^x) = 3/(3+9^x)</p><p><strong>Step 2:</strong> Add the two fractions on RHS.</p><p>RHS = 9^x/(9^x+3) + 3/(9^x+3) = (9^x+3)/(9^x+3) = 1</p><p>Therefore: <strong>f(x) + f(1-x) = 1</strong></p><p><strong>Step 3:</strong> Use symmetry to pair terms in the sum.</p><p>For the sum ∑(r=1 to 2008) f(r/2009), 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 4:</strong> Count the pairs.</p><p>The terms are: f(1/2009), f(2/2009), ..., f(2008/2009)</p><p>We can pair: (r=1 with r=2008), (r=2 with r=2007), ..., (r=1004 with r=1005)</p><p>Number of pairs = 2008/2 = 1004</p><p>Each pair sums to 1.</p><p><strong>∴ Answer: 1004</strong>
Correct Answer: 1004