Let $$\mathbb{R}$$ denote the set of all real numbers. Let $$f: \mathbb{R} \to \mathbb{R}$$ be a function such that $$f(x) > 0$$ for all $$x \in \mathbb{R}$$, and $$f(x + y) = f(x)f(y)$$ for all $$x, y \in \mathbb{R}$$.

Let the real numbers $$a_1, a_2, \ldots, a_{50}$$ be in an arithmetic progression. If $$f(a_{31}) = 64f(a_{25})$$, and

$$\sum_{i=1}^{50} f(a_i) = 3(2^{25} + 1),$$

then the value of

$$\sum_{i=6}^{30} f(a_i)$$

is ________.

The functional equation $$f(x+y)=f(x)f(y)$$ with the extra condition $$f(x)\gt 0$$ for all real $$x$$ has the well-known positive solutions
    $$f(x)=b^{x}$$ for some constant $$b\gt 0$$.

Let the arithmetic progression be
    $$a_i=a_1+(i-1)d,\qquad i=1,2,\dots ,50.$$

Then the corresponding function values form a geometric progression:

$$f(a_i)=b^{a_i}=b^{\,a_1+(i-1)d}=b^{a_1}\bigl(b^{d}\bigr)^{\,i-1}.$$

Set $$r=b^{d}\;( \gt 0).$$  So
    $$f(a_i)=f(a_1)\,r^{\,i-1}\qquad (1).$$

Step 1: Find the common ratio $$r$$.

Given $$f(a_{31})=64\,f(a_{25}).$$ Using (1):

$$\frac{f(a_{31})}{f(a_{25})}=r^{31-25}=r^{6}=64=2^{6}\; \Longrightarrow\; r=2.$$(2)

Step 2: Determine $$f(a_1).$$

The first 50 terms sum to $$3\bigl(2^{25}+1\bigr):$$

Using (1) with $$r=2$$,

$$\sum_{i=1}^{50} f(a_i)=f(a_1)\sum_{i=0}^{49} 2^{\,i}=f(a_1)\,\frac{2^{50}-1}{2-1} =f(a_1)\,\bigl(2^{50}-1\bigr).$$

Equate this to the given value:

$$f(a_1)\,\bigl(2^{50}-1\bigr)=3\bigl(2^{25}+1\bigr).$$

Write $$A=2^{25}.$$ Then $$2^{50}=A^{2}$$, and

$$f(a_1)\,(A^{2}-1)=3(A+1).$$

Because $$A^{2}-1=(A-1)(A+1),$$ we get

$$f(a_1)=\frac{3(A+1)}{(A-1)(A+1)}=\frac{3}{A-1}=\frac{3}{2^{25}-1}.\qquad(3)$$

Step 3: Sum from $$i=6$$ to $$i=30$$.

This sum contains $$30-6+1=25$$ terms. The first term in this part is $$f(a_6)=f(a_1)\,2^{\,5}$$. Hence

$$\sum_{i=6}^{30} f(a_i)=f(a_1)\,2^{5}\sum_{k=0}^{24} 2^{\,k} =f(a_1)\,32\;\frac{2^{25}-1}{2-1} =32\,f(a_1)\,\bigl(2^{25}-1\bigr).$$

Using (3), $$f(a_1)\,\bigl(2^{25}-1\bigr)=3,$$ so

$$\sum_{i=6}^{30} f(a_i)=32\times 3=96.$$

Therefore, the required value is 96.