Each pair sums to (1)/(2). There are 29 such pairs. The middle term is f(2) = (2)/(2² + 4) = (2)/(4+4) = (2)/(8) = (1)/(4). So the total sum is 29 · (1)/(2) + (1)/(4) = (58)/(4) + (1)/(4) = (59)/(4).

Each pair sums to (1)/(2). There are 29 such pairs. The middle term is f(2) = (2)/(2² + 4) = (2)/(4+4) = (2)/(8) = (1)/(4). So the total sum is 29 · (1)/(2) + (1)/(4) = (58)/(4) + (1)/(4) = (59)/(4).

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

Answer: 118

Step-by-step solution

Step 1: Simplify the function $f(x)$

First, we simplify the given function $f(x)$ . We can rewrite the terms in the numerator and denominator using the exponent rule $a^{m+n} = a^m \cdot a^n$ . This allows us to express all terms in terms of $2^x$ .

Step 2: Factor out common terms

Next, we factor out common terms from the numerator and denominator. In the numerator, we factor out 4. In the denominator, we factor out 2 and recognize that the remaining quadratic expression is a perfect square trinomial, $(2^x + 4)^2$ .

Step 3: Further simplify $f(x)$

We can cancel out one factor of $(2^x + 4)$ from the numerator and denominator, simplifying the function to $f(x) = \frac{2}{2^x + 4}$ . This is the most simplified form of the function.

Step 4: Find the relationship between $f(x)$ and $f(4-x)$

Now, let's evaluate $f(4-x)$ to see if there's a useful relationship. We substitute $(4-x)$ into the simplified function. After algebraic manipulation, we find that $f(x) + f(4-x) = \frac{2}{2^x + 4} + \frac{2^x}{8 + 2 \cdot 2^x} = \frac{2}{2^x + 4} + \frac{2^x}{2(4 + 2^x)} = \frac{4 + 2^x}{2(2^x + 4)} = \frac{1}{2}$ .

Step 5: Pair the terms in the sum

The sum is $S = f\left(\frac{1}{15}\right) + f\left(\frac{2}{15}\right) + \ldots + f\left(\frac{59}{15}\right)$ . We notice that the terms can be paired such that the sum of the arguments is 4. For example, $\frac{1}{15} + \frac{59}{15} = \frac{60}{15} = 4$ . There are 29 such pairs, and one middle term $f\left(\frac{30}{15}\right) = f(2)$ .

Step 6: Calculate the sum

Each pair sums to $\frac{1}{2}$ . There are 29 such pairs. The middle term is $f(2) = \frac{2}{2^2 + 4} = \frac{2}{4+4} = \frac{2}{8} = \frac{1}{4}$ . So the total sum is $29 \cdot \frac{1}{2} + \frac{1}{4} = \frac{58}{4} + \frac{1}{4} = \frac{59}{4}$ .

Step 7: Calculate the final value

Finally, we need to multiply the sum $S$ by 8 to get the required value.

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