[THOUGHT] We are given a function $f: \mathbb{R} \to \mathbb{R}$ such that $f(x) > 0$ for all $x$, and $f(x+y) = f(x)f(y)$ for all $x, y$. This is a Cauchy exponential functional equation. Under positivity, the solution is $f(x) = b^x$ for some $b > 0$. Let $f(x) = r^x$ with $r > 0$.

We have arithmetic progression $a_1, a_2, \dots, a_{50}$. Let the common difference be $d$, and first term $a_1 = A$. Then $a_k = A + (k-1)d$.

Given $f(a_{31}) = 64 f(a_{25})$. Since $f(a_k) = r^{a_k}$, we have:

Now $a_{31} - a_{25} = (A + 30d) - (A + 24d) = 6d$. So:

Thus $r^{6d} = 2^6 \implies (r^d)^6 = 2^6 \implies r^d = 2$ (since $r > 0$, $r^d > 0$, take positive 6th root). So $r^d = 2$.

Now we are given:

We have $f(a_i) = r^{a_i} = r^{A + (i-1)d} = r^A \cdot (r^d)^{i-1} = r^A \cdot 2^{i-1}$.

Thus:

Given this equals $3(2^{25} + 1)$. So:

Note $2^{50} - 1 = (2^{25} - 1)(2^{25} + 1)$. Therefore:

Thus $r^A = \frac{3}{2^{25} - 1}$.

Now we need $\sum_{i=6}^{30} f(a_i)$. This is:

Compute the sum $S = \sum_{i=6}^{30} 2^{i-1} = \sum_{j=5}^{29} 2^{j}$ (since $i-1 = j$, when $i=6, j=5$, when $i=30, j=29$). So:

This is a geometric series with first term $2^5 = 32$, ratio 2, number of terms = $29 - 5 + 1 = 25$. Sum = $32 \cdot \frac{2^{25} - 1}{2 - 1} = 32 (2^{25} - 1)$.

Thus:

So the answer is 96.

Let's verify with Python to ensure no algebraic mistake.

We'll compute numerically: choose $d = 1$, then $r^d = 2 \implies r = 2$. Then $r^A = \frac{3}{2^{25} - 1} \implies 2^A = \frac{3}{2^{25} - 1} \implies A = \log_2(3/(2^{25}-1))$. Then compute $f(a_i) = 2^{A + (i-1)}$. Sum from i=1 to 50 should equal $3(2^{25}+1)$. Then sum from i=6 to 30 should be 96.

We'll do that in Python. [/THOUGHT] [PYTHON] import sympy as sp

Let d = 1, r = 2

d = 1 r = 2

Compute A such that r^A * (2^50 - 1) = 3*(2^25 + 1)

r^A = 3/(2^25 - 1)

A_val = sp.log(3/(2**25 - 1), 2) # log base 2

Now compute f(a_i) = r^(A + (i-1)*d) = 2^(A + i - 1)

Sum i=1 to 50

total_sum = 0 for i in range(1, 51): term = 2**(A_val + (i-1)) total_sum += term

Convert to exact rational maybe? Use sympy to simplify.

total_sum_simplified = sp.simplify(total_sum) print("Total sum (i=1 to 50):", total_sum_simplified) print("Expected total sum:", 3*(2**25 + 1))

Sum i=6 to 30

sum_6_to_30 = 0 for i in range(6, 31): term = 2**(A_val + (i-1)) sum_6_to_30 += term

print("Sum i=6 to 30:", sum_6_to_30) print("Simplified sum:", sp.simplify(sum_6_to_30)) print("Expected sum: 96")

Check if sum_6_to_30 equals 96

print("Is sum_6_to_30 == 96?", sp.simplify(sum_6_to_30 - 96) == 0) [/PYTHON]

[THOUGHT] The Python output confirms that the sum from i=6 to 30 simplifies to 96 exactly. The earlier reasoning gave 96, and the numerical computation supports it. The answer is 96.

$\boxed{96}$