Key Concepts and Formulas

• Algebraic Simplification: Manipulating algebraic expressions to simplify them. This includes factoring, substitution, and recognizing common patterns.
• Properties of Exponents: Rules such as $a^{m+n} = a^m \cdot a^n$ and $(a^m)^n = a^{mn}$.
• Symmetry Property for Series: If a function $f(x)$ satisfies $f(x) + f(a-x) = C$ for some constant $C$, then the sum of $f(x_i)$ where $x_i$ are in an arithmetic progression can be simplified by pairing terms.

Step-by-Step Solution

Step 1: Simplify the function $f(x)$. Let $y = 2^x$. Then the function can be rewritten in terms of $y$. The numerator is $2^{x+2} + 16 = 2^x \cdot 2^2 + 16 = 4 \cdot 2^x + 16 = 4y + 16$. The denominator is $2^{2x+1} + 2^{x+4} + 32 = 2^{2x} \cdot 2^1 + 2^x \cdot 2^4 + 32 = 2 \cdot (2^x)^2 + 16 \cdot 2^x + 32 = 2y^2 + 16y + 32$. So, $f(x) = \frac{4y + 16}{2y^2 + 16y + 32}$.

Step 2: Further simplify the expression for $f(x)$. We can factor out constants from the numerator and the denominator. Numerator: $4y + 16 = 4(y + 4)$. Denominator: $2y^2 + 16y + 32 = 2(y^2 + 8y + 16) = 2(y+4)^2$. Therefore, $f(x) = \frac{4(y+4)}{2(y+4)^2} = \frac{2}{y+4}$. Substituting back $y = 2^x$, we get $f(x) = \frac{2}{2^x + 4}$.

Step 3: Analyze the sum and identify a symmetry property. We need to calculate $S = f\left(\frac{1}{15}\right)+f\left(\frac{2}{15}\right)+\ldots+f\left(\frac{59}{15}\right)$. Let's check if there is a symmetry property of the form $f(x) + f(a-x) = C$. Consider $f(x) + f\left(\frac{60}{15}-x\right) = f(x) + f(4-x)$. $f(x) = \frac{2}{2^x + 4}$. $f(4-x) = \frac{2}{2^{4-x} + 4} = \frac{2}{\frac{2^4}{2^x} + 4} = \frac{2}{\frac{16}{2^x} + 4} = \frac{2 \cdot 2^x}{16 + 4 \cdot 2^x} = \frac{2 \cdot 2^x}{4(4 + 2^x)} = \frac{2^x}{2(4 + 2^x)}$. Now, let's sum $f(x)$ and $f(4-x)$: $f(x) + f(4-x) = \frac{2}{2^x + 4} + \frac{2^x}{2(4 + 2^x)} = \frac{2 \cdot 2}{2(2^x + 4)} + \frac{2^x}{2(4 + 2^x)} = \frac{4 + 2^x}{2(2^x + 4)} = \frac{1}{2}$. So, the symmetry property is $f(x) + f(4-x) = \frac{1}{2}$.

Step 4: Apply the symmetry property to the sum. The sum is $S = f\left(\frac{1}{15}\right)+f\left(\frac{2}{15}\right)+\ldots+f\left(\frac{59}{15}\right)$. The terms in the sum are of the form $f(x_k)$ where $x_k = \frac{k}{15}$ for $k = 1, 2, \ldots, 59$. The total number of terms is 59. We can pair terms such that $x_k + x_{60-k} = \frac{k}{15} + \frac{60-k}{15} = \frac{60}{15} = 4$. So, $f(x_k) + f(x_{60-k}) = f\left(\frac{k}{15}\right) + f\left(\frac{60-k}{15}\right) = f\left(\frac{k}{15}\right) + f\left(4 - \frac{k}{15}\right) = \frac{1}{2}$.

Let's consider the pairing for the sum $S$: The terms are $f\left(\frac{1}{15}\right), f\left(\frac{2}{15}\right), \ldots, f\left(\frac{59}{15}\right)$. We can pair $f\left(\frac{1}{15}\right)$ with $f\left(\frac{59}{15}\right)$, since $\frac{1}{15} + \frac{59}{15} = \frac{60}{15} = 4$. Their sum is $\frac{1}{2}$. We can pair $f\left(\frac{2}{15}\right)$ with $f\left(\frac{58}{15}\right)$, since $\frac{2}{15} + \frac{58}{15} = \frac{60}{15} = 4$. Their sum is $\frac{1}{2}$. This pairing continues up to $f\left(\frac{29}{15}\right)$ and $f\left(\frac{31}{15}\right)$. The middle term is $f\left(\frac{30}{15}\right) = f(2)$.

The sum can be written as: $S = \left[f\left(\frac{1}{15}\right) + f\left(\frac{59}{15}\right)\right] + \left[f\left(\frac{2}{15}\right) + f\left(\frac{58}{15}\right)\right] + \ldots + \left[f\left(\frac{29}{15}\right) + f\left(\frac{31}{15}\right)\right] + f\left(\frac{30}{15}\right)$. There are 29 such pairs, and each pair sums to $\frac{1}{2}$. So, the sum of the pairs is $29 \times \frac{1}{2} = \frac{29}{2}$.

Now, we need to calculate the middle term $f\left(\frac{30}{15}\right) = f(2)$. Using the simplified function $f(x) = \frac{2}{2^x + 4}$: $f(2) = \frac{2}{2^2 + 4} = \frac{2}{4 + 4} = \frac{2}{8} = \frac{1}{4}$.

The total sum $S$ is the sum of the pairs plus the middle term: $S = \frac{29}{2} + \frac{1}{4} = \frac{58}{4} + \frac{1}{4} = \frac{59}{4}$.

Step 5: Calculate the final required value. We are asked to find the value of $8 \times S$. $8 \times S = 8 \times \frac{59}{4} = 2 \times 59 = 118$.

Common Mistakes & Tips

• Algebraic Errors: Be very careful with algebraic manipulations, especially with exponents. Double-check factoring and substitutions.
• Identifying the Symmetry: The key to solving this problem efficiently is recognizing the symmetry property. If you miss it, the direct calculation of 59 terms will be extremely tedious and prone to errors.
• Handling the Middle Term: In sums with an odd number of terms, always ensure the middle term is correctly identified and added separately after pairing.

Summary

The problem involves evaluating a sum of a function $f(x)$ over a range of arguments. The first step is to simplify the function $f(x)$ algebraically. Upon simplification, we found $f(x) = \frac{2}{2^x + 4}$. Then, we discovered a crucial symmetry property: $f(x) + f(4-x) = \frac{1}{2}$. This property allows us to pair terms in the given sum. The sum consists of 59 terms, which can be grouped into 29 pairs, each summing to $\frac{1}{2}$, and one middle term, $f(2)$, which equals $\frac{1}{4}$. The total sum is $\frac{29}{2} + \frac{1}{4} = \frac{59}{4}$. Finally, we multiply this sum by 8, yielding $8 \times \frac{59}{4} = 118$.

The final answer is \boxed{118}.