[THOUGHT] The problem asks for the sum of a specific range of terms in a sequence generated by a function $f$ applied to an arithmetic progression. The function $f$ satisfies $f(x+y) = f(x)f(y)$, which implies $f(x)$ is exponential, so $f(a_i)$ forms a geometric progression. I derived that the common ratio $R=2$ and the first term $A = 3/(2^{25}-1)$. The target sum is $\sum_{i=6}^{30} f(a_i)$. I will use Python to verify the algebraic steps and compute the final value. [/THOUGHT] [PYTHON] from sympy import symbols, solve, Sum

Define variables

A, R = symbols('A R', real=True, positive=True)

Condition 1: f(a_31) = 64 * f(a_25)

f(a_i) corresponds to the i-th term of a geometric sequence with first term A and ratio R.

The term at index i is A * R**(i-1).

So f(a_31) is A * R30 and f(a_25) is A * R24.

eq1 = A * R30 - 64 * A * R24

Solve for R

sol_R = solve(eq1, R) print(f"Possible values for R: {sol_R}")

We know f(x) > 0, so R must be positive.

From the output, we expect R=2.

R_val = 2

Condition 2: Sum_{i=1}^{50} f(a_i) = 3 * (2**25 + 1)

Sum of geometric series: A * (R**n - 1) / (R - 1)

sum_50 = A * (R_val50 - 1) / (R_val - 1) given_sum = 3 * (225 + 1) eq2 = sum_50 - given_sum sol_A = solve(eq2, A) A_val = sol_A[0] print(f"Value of A: {A_val}")

Target: Sum_{i=6}^{30} f(a_i)

This is a sum of a geometric series starting at term 6 (A * R**5) with 25 terms (30 - 6 + 1).

Formula: first_term * (R**count - 1) / (R - 1)

first_term_target = A_val * R_val5 count_target = 30 - 6 + 1 target_sum = first_term_target * (R_valcount_target - 1) / (R_val - 1)

print(f"Target Sum expression: {target_sum}") print(f"Target Sum value: {target_sum.evalf()}") [/PYTHON] [THOUGHT] The Python code confirmed the algebraic derivation. $R = 2$. $A = 3 / (2^{25} - 1)$. The target sum evaluates to 96. I am confident in this result. [/THOUGHT] \boxed{96}

The function $f$ satisfies $f(x+y) = f(x)f(y)$ and $f(x) > 0$, which implies $f(x) = c^x$ for some constant $c > 0$. Since $a_1, a_2, \ldots, a_{50}$ is an arithmetic progression with common difference $d$, the sequence $f(a_1), f(a_2), \ldots, f(a_{50})$ is a geometric progression. Let $r = f(d)$ be the common ratio and $A = f(a_1)$ be the first term. Then $f(a_i) = A r^{i-1}$.

Given $f(a_{31}) = 64 f(a_{25})$, we have: [ A r^{30} = 64 A r^{24} \implies r^6 = 64 \implies r = 2 ] (since $f(x) > 0$, the ratio must be positive).

Given $\sum_{i=1}^{50} f(a_i) = 3(2^{25} + 1)$, the sum of the first 50 terms of the geometric progression is: [ \frac{A(r^{50} - 1)}{r - 1} = 3(2^{25} + 1) ] Substituting $r=2$: [ A(2^{50} - 1) = 3(2^{25} + 1) ] Using the difference of squares $2^{50} - 1 = (2^{25} - 1)(2^{25} + 1)$: [ A(2^{25} - 1)(2^{25} + 1) = 3(2^{25} + 1) ] [ A(2^{25} - 1) = 3 \implies A = \frac{3}{2^{25} - 1} ]

We want to find $\sum_{i=6}^{30} f(a_i)$. This is the sum of terms from index 6 to 30 (inclusive), which has $30 - 6 + 1 = 25$ terms. The first term in this partial sum is $f(a_6) = A r^5$. [ \sum_{i=6}^{30} f(a_i) = \frac{f(a_6)(r^{25} - 1)}{r - 1} = \frac{A r^5 (r^{25} - 1)}{r - 1} ] Substituting $r=2$ and $A = \frac{3}{2^{25} - 1}$: [ \text{Sum} = \frac{\frac{3}{2^{25} - 1} \cdot 2^5 \cdot (2^{25} - 1)}{2 - 1} ] [ \text{Sum} = 3 \cdot 2^5 = 3 \cdot 32 = 96 ]

\boxed{96}