Question

Let \( f(x) = \frac{2^{x+2} + 16}{2^{2x+1} + 2^{x+4} + 32} \). Then the value of \[ 8 \left( f\left( \frac{1}{15} \right) + f\left( \frac{2}{15} \right) + \dots + f\left( \frac{59}{15} \right) \right) \] is equal to:

• A. 118
• B. 92
• C. 102
• D. 108

Hint:

For complex sums involving functions of fractions, try to simplify the function first and look for any symmetrical or repetitive patterns in the terms. Sometimes numerical evaluation can provide quick results.

Answer 1

The Correct Option is A

Step 1: Simplify the expression for f(x). Given \( f(x) = \frac{2^{x+2} + 16}{2^{2x+1} + 2^{x+4} + 32} \). 
We can rewrite this as: \[ f(x) = \frac{2^2 \cdot 2^x + 16}{2 \cdot (2^x)^2 + 2^4 \cdot 2^x + 32} = \frac{4 \cdot 2^x + 16}{2 \cdot (2^x)^2 + 16 \cdot 2^x + 32}. \] Factoring out constants, we get: \[ f(x) = \frac{4(2^x + 4)}{2((2^x)^2 + 8 \cdot 2^x + 16)} = \frac{2(2^x + 4)}{(2^x + 4)^2} = \frac{2}{2^x + 4}. \] Step 2: Determine the pairing property. Let's examine \( f(a) + f(4-a) \): \[ f(a) + f(4-a) = \frac{2}{2^a + 4} + \frac{2}{2^{4-a} + 4} = \frac{2}{2^a + 4} + \frac{2}{\frac{16}{2^a} + 4}. \] Multiplying the numerator and denominator of the second term by \( 2^a \), we have: \[ f(a) + f(4-a) = \frac{2}{2^a + 4} + \frac{2 \cdot 2^a}{16 + 4 \cdot 2^a} = \frac{2}{2^a + 4} + \frac{2^a}{8 + 2 \cdot 2^a} = \frac{2}{2^a + 4} + \frac{2^a}{2(4 + 2^a)}. \] Combining the fractions: \[ f(a) + f(4-a) = \frac{4 + 2^a}{2(2^a + 4)} = \frac{1}{2}. \] Step 3: Evaluate the sum. 
We need to calculate \( 8 \left( \sum_{k=1}^{59} f\left(\frac{k}{15}\right) \right) \). 
We pair terms of the form \( \frac{k}{15} \) and \( 4 - \frac{k}{15} \). Notice that \( 4 - \frac{k}{15} = \frac{60-k}{15} \). 
We can rewrite the sum as: \[ \sum_{k=1}^{59} f\left(\frac{k}{15}\right) = \sum_{k=1}^{29} \left[ f\left(\frac{k}{15}\right) + f\left(\frac{60-k}{15}\right) \right] + f\left(\frac{30}{15}\right) = \sum_{k=1}^{29} \left[ f\left(\frac{k}{15}\right) + f\left(4 - \frac{k}{15}\right) \right] + f(2). \] Using the result from Step 2: \[ \sum_{k=1}^{59} f\left(\frac{k}{15}\right) = \sum_{k=1}^{29} \frac{1}{2} + f(2) = 29 \cdot \frac{1}{2} + \frac{2}{2^2 + 4} = \frac{29}{2} + \frac{2}{8} = \frac{29}{2} + \frac{1}{4} = \frac{58+1}{4} = \frac{59}{4}. \] Step 4: Calculate the final result. Finally, \[ 8 \left( \sum_{k=1}^{59} f\left(\frac{k}{15}\right) \right) = 8 \cdot \frac{59}{4} = 2 \cdot 59 = 118. \] Therefore, the answer is 118.