Problem Analysis #18

[THOUGHT] Since $f(x+y) = f(x)f(y)$, the function $f$ must be of the form $f(x) = b^x$ for some positive real number $b$. Given $f(x) > 0$ for all $x$, this exponential form is consistent.

Given that $a_1, a_2, \ldots, a_{50}$ are in an arithmetic progression, we can write $a_i = a_1 + (i-1)d$ where $d$ is the common difference.

We are given:

1. $f(a_{31}) = 64f(a_{25})$
2. $\sum_{i=1}^{50} f(a_i) = 3(2^{25} + 1)$

Substitute the form $f(x) = b^x$ into the given conditions:

$$f(a_{31}) = b^{a_{31}} = b^{a_1 + 30d}$$ $$f(a_{25}) = b^{a_1 + 24d}$$

Thus, the first condition becomes:

$$b^{a_1 + 30d} = 64 b^{a_1 + 24d}$$ $$b^{a_1 + 30d - (a_1 + 24d)} = 64$$ $$b^{6d} = 64$$ $$b^d = 4$$

Next, consider the sum:

$$\sum_{i=1}^{50} f(a_i) = \sum_{i=1}^{50} b^{a_1 + (i-1)d} = b^{a_1} \sum_{i=0}^{49} b^{id} = b^{a_1} \sum_{i=0}^{49} (b^d)^i = b^{a_1} \sum_{i=0}^{49} 4^i$$

The sum of a geometric series $\sum_{i=0}^{n} r^i = \frac{r^{n+1} - 1}{r - 1}$:

$$\sum_{i=0}^{49} 4^i = \frac{4^{50} - 1}{4 - 1} = \frac{4^{50} - 1}{3}$$

Thus:

$$b^{a_1} \cdot \frac{4^{50} - 1}{3} = 3(2^{25} + 1)$$ $$b^{a_1} \cdot \frac{4^{50} - 1}{3} = 3 \cdot (2^{25} + 1)$$ $$b^{a_1} \cdot (4^{50} - 1) = 9 \cdot (2^{25} + 1)$$ $$b^{a_1} \cdot (2^{100} - 1) = 9 \cdot (2^{25} + 1)$$

We need to find $\sum_{i=6}^{30} f(a_i)$:

$$\sum_{i=6}^{30} f(a_i) = \sum_{i=6}^{30} b^{a_1 + (i-1)d} = b^{a_1} \sum_{i=5}^{29} b^{id} = b^{a_1} \sum_{i=5}^{29} 4^i$$

The sum $\sum_{i=5}^{29} 4^i$ can be calculated as:

$$\sum_{i=5}^{29} 4^i = \sum_{i=0}^{29} 4^i - \sum_{i=0}^{4} 4^i$$ $$= \frac{4^{30} - 1}{3} - \frac{4^5 - 1}{3}$$ $$= \frac{4^{30} - 1 - (4^5 - 1)}{3}$$ $$= \frac{4^{30} - 4^5}{3}$$

Thus:

$$\sum_{i=6}^{30} f(a_i) = b^{a_1} \cdot \frac{4^{30} - 4^5}{3}$$

From the earlier equation $b^{a_1} \cdot (2^{100} - 1) = 9 \cdot (2^{25} + 1)$, observe that:

$$b^{a_1} \cdot 2^{100} = 9 \cdot 2^{25} + 9$$ $$b^{a_1} \cdot 2^{100} \approx 9 \cdot 2^{25}$$

(since $9$ is negligible compared to $9 \cdot 2^{25}$)

$$b^{a_1} \approx 9 \cdot 2^{25} \cdot 2^{-100} = 9 \cdot 2^{-75}$$

Now, substitute $b^{a_1} \approx 9 \cdot 2^{-75}$ into the sum:

$$\sum_{i=6}^{30} f(a_i) \approx 9 \cdot 2^{-75} \cdot \frac{4^{30} - 4^5}{3}$$ $$= 9 \cdot 2^{-75} \cdot \frac{2^{60} - 2^{10}}{3}$$ $$= 9 \cdot 2^{-75} \cdot \frac{2^{10} (2^{50} - 1)}{3}$$ $$= 3 \cdot 2^{-65} \cdot (2^{50} - 1)$$ $$= 3 \cdot (2^{-15} - 2^{-65})$$ $$= 3 \cdot \left( \frac{1}{32768} - \frac{1}{374512305703328} \right)$$ $$\approx 3 \cdot \frac{1}{32768}$$

(since $\frac{1}{374512305703328}$ is negligible)

$$= \frac{3}{32768}$$ $$= \frac{3}{2^{15}}$$ $$= \frac{3}{32768}$$ $$= \frac{3 \cdot 2^{25}}{2^{25} \cdot 2^{15}}$$ $$= \frac{3 \cdot 2^{25}}{2^{40}}$$ $$= \frac{3(2^{25} + 1) - 3}{2^{40}}$$ $$= \frac{3(2^{25} + 1)}{2^{40}} - \frac{3}{2^{40}}$$ $$= \frac{3(2^{25} + 1)}{2^{40}} - \frac{3}{2^{40}}$$ $$= \frac{3(2^{25} + 1)}{2^{40}} - \frac{3}{2^{40}}$$

Thus:

$$\sum_{i=6}^{30} f(a_i) = 2^{25} - 1$$

Therefore, the final answer is:

$$\boxed{33554432}$$