Problem Analysis #18

[THOUGHT] The functional equation $f(x+y) = f(x)f(y)$ with $f(x) > 0$ implies $f(x) = a^x$ for some $a > 0$. Let the arithmetic progression be $a_i = a + (i-1)d$. Given $f(a_{31}) = 64f(a_{25})$, we derive $a^d = 2$. The sum $\sum_{i=1}^{50} f(a_i) = C \cdot \frac{2^{50} - 1}{2 - 1} = 3(2^{25} + 1)$, where $C = a^a$. This simplifies to $C(2^{50} - 1) = 3(2^{25} + 1)$. The desired sum $\sum_{i=6}^{30} f(a_i) = C \cdot 2^5 \cdot \frac{2^{25} - 1}{2 - 1} = 32C(2^{25} - 1)$. Using $C(2^{50} - 1) = 3(2^{25} + 1)$ and $2^{50} - 1 = (2^{25} - 1)(2^{25} + 1)$, we find $C(2^{25} - 1) = 3$, so the sum equals $32 \cdot 3 = 96$.

[/THOUGHT] [PYTHON] t = 2**25 total_sum = 3*(t+1) partial_sum = total_sum * 32 / (t+1) print(partial_sum) [/PYTHON]

[THOUGHT] The Python code execution confirms the derived result. The sum $\sum_{i=6}^{30} f(a_i)$ is calculated as 96.0, which is an integer and matches our analytical solution.

[/THOUGHT] \boxed{96}