[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: [ r^{a_{31}} = 64 , r^{a_{25}} \implies r^{a_{31} - a_{25}} = 64. ] Now ( a_{31} - a_{25} = (A + 30d) - (A + 24d) = 6d ). So: [ r^{6d} = 64 = 2^6. ] 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: [ \sum_{i=1}^{50} f(a_i) = 3(2^{25} + 1). ] 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: [ \sum_{i=1}^{50} f(a_i) = r^A \sum_{i=1}^{50} 2^{i-1} = r^A \cdot \frac{2^{50} - 1}{2 - 1} = r^A (2^{50} - 1). ] Given this equals ( 3(2^{25} + 1) ). So: [ r^A (2^{50} - 1) = 3(2^{25} + 1). ] Note ( 2^{50} - 1 = (2^{25} - 1)(2^{25} + 1) ). Therefore: [ r^A (2^{25} - 1)(2^{25} + 1) = 3(2^{25} + 1) \implies r^A (2^{25} - 1) = 3. ] Thus ( r^A = \frac{3}{2^{25} - 1} ).

Now we need ( \sum_{i=6}^{30} f(a_i) ). This is: [ \sum_{i=6}^{30} f(a_i) = \sum_{i=6}^{30} r^{a_i} = \sum_{i=6}^{30} r^A \cdot 2^{i-1} = r^A \sum_{i=6}^{30} 2^{i-1}. ] 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: [ S = \sum_{j=5}^{29} 2^{j} = 2^{5} + 2^{6} + \dots + 2^{29}. ] 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: [ \sum_{i=6}^{30} f(a_i) = r^A \cdot 32 (2^{25} - 1) = \frac{3}{2^{25} - 1} \cdot 32 (2^{25} - 1) = 3 \cdot 32 = 96. ]

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}