Step-by-step Solution:

1. Determine the form of the function f(x).

The function f: ℝ → ℝ satisfies the following properties:

• f(x) > 0 for all x ∈ ℝ.
• f(x+y) = f(x)f(y) for all x, y ∈ ℝ.

This is a standard functional equation. The solution is an exponential function of the form $f(x) = b^x$ for some constant b > 0.

2. Analyze the sequence f(aᵢ).

The numbers a₁, a₂, ..., a₅₀ are in an arithmetic progression (AP). Let the first term be a and the common difference be d. Then the i-th term is given by aᵢ = a + (i-1)d.

Now, let's consider the sequence f(aᵢ): $f(a_i) = b^{a_i} = b^{a + (i-1)d} = b^a \cdot (b^d)^{i-1}$ This shows that the sequence f(a₁), f(a₂), ... is a geometric progression (GP). Let the first term of this GP be $A = f(a₁) = b^a$ and the common ratio be $R = b^d$. So, $f(aᵢ) = A \cdot R^{i-1}$.

3. Find the common ratio R.

We are given the condition f(a₃₁) = 64f(a₂₅). Using the formula for the terms of the GP:

• $f(a₃₁) = A \cdot R^{31-1} = A \cdot R^{30}$
• $f(a₂₅) = A \cdot R^{25-1} = A \cdot R^{24}$

Substituting these into the given condition: $A \cdot R^{30} = 64 \cdot (A \cdot R^{24})$ Since A = f(a₁) > 0, we can divide both sides by A: $R^{30} = 64 R^{24}$ $R^{30} / R^{24} = 64$ $R^6 = 64 = 2^6$ Since $R = b^d$ and b > 0, R must be positive. Therefore, R = 2.

4. Use the sum of the first 50 terms to find an expression involving A.

We are given that $\sum\limits_{i=1}^{50} f(a_i) = 3(2^{25}+1)$. This is the sum of the first 50 terms of the GP with first term A and common ratio R=2. The sum of the first n terms of a GP is $S_n = A(R^n - 1) / (R - 1)$.

For n=50, R=2: $S_{50} = \frac{A(2^{50} - 1)}{2 - 1} = A(2^{50} - 1)$ Equating this to the given value: $A(2^{50} - 1) = 3(2^{25}+1)$ We can factor $2^{50} - 1$ as a difference of squares: $2^{50} - 1 = (2^{25})^2 - 1^2 = (2^{25} - 1)(2^{25} + 1)$. $A(2^{25} - 1)(2^{25} + 1) = 3(2^{25}+1)$ Since $2^{25} + 1 ≠ 0$, we can divide both sides by it: $A(2^{25} - 1) = 3$ We will use this result in the next step.

5. Calculate the required sum.

We need to find the value of $\sum\limits_{i=6}^{30} f(a_i)$. This is the sum of terms of the GP from the 6th term to the 30th term. This is itself a GP with:

• First term: $f(a₆) = A \cdot R^{6-1} = A \cdot 2^5 = 32A$.
• Number of terms: 30 - 6 + 1 = 25.
• Common ratio: R = 2.

The sum S' of this series is: $S' = \frac{(\text{first term}) \cdot (R^{\text{number of terms}} - 1)}{R - 1}$ $S' = \frac{32A \cdot (2^{25} - 1)}{2 - 1}$ $S' = 32 \cdot A(2^{25} - 1)$ From Step 4, we know that $A(2^{25} - 1) = 3$. Substituting this value into the expression for S': $S' = 32 \cdot 3 = 96$

Thus, the value of $\sum\limits_{i=6}^{30} f(a_i)$ is 96.