Problem Analysis #18

[THOUGHT] Since ( f(x+y) = f(x)f(y) ), the function ( f ) must be of the form ( f(x) = b^x ) for some positive real number ( b ). Given ( f(x) > 0 ) for all ( x ), this exponential form is consistent.

Given that ( a_1, a_2, \ldots, a_{50} ) are in an arithmetic progression, we can write ( a_i = a_1 + (i-1)d ) where ( d ) is the common difference.

We are given:

1. ( f(a_{31}) = 64f(a_{25}) )
2. ( \sum_{i=1}^{50} f(a_i) = 3(2^{25} + 1) )

Substitute the form ( f(x) = b^x ) into the given conditions: [ f(a_{31}) = b^{a_{31}} = b^{a_1 + 30d} ] [ f(a_{25}) = b^{a_1 + 24d} ]

Thus, the first condition becomes: [ b^{a_1 + 30d} = 64 b^{a_1 + 24d} ] [ b^{a_1 + 30d - (a_1 + 24d)} = 64 ] [ b^{6d} = 64 ] [ b^d = 4 ]

Next, consider the sum: [ \sum_{i=1}^{50} f(a_i) = \sum_{i=1}^{50} b^{a_1 + (i-1)d} = b^{a_1} \sum_{i=0}^{49} b^{id} = b^{a_1} \sum_{i=0}^{49} (b^d)^i = b^{a_1} \sum_{i=0}^{49} 4^i ]

The sum of a geometric series ( \sum_{i=0}^{n} r^i = \frac{r^{n+1} - 1}{r - 1} ): [ \sum_{i=0}^{49} 4^i = \frac{4^{50} - 1}{4 - 1} = \frac{4^{50} - 1}{3} ]

Thus: [ b^{a_1} \cdot \frac{4^{50} - 1}{3} = 3(2^{25} + 1) ] [ b^{a_1} \cdot \frac{4^{50} - 1}{3} = 3 \cdot (2^{25} + 1) ] [ b^{a_1} \cdot (4^{50} - 1) = 9 \cdot (2^{25} + 1) ] [ b^{a_1} \cdot (2^{100} - 1) = 9 \cdot (2^{25} + 1) ]

We need to find ( \sum_{i=6}^{30} f(a_i) ): [ \sum_{i=6}^{30} f(a_i) = \sum_{i=6}^{30} b^{a_1 + (i-1)d} = b^{a_1} \sum_{i=5}^{29} b^{id} = b^{a_1} \sum_{i=5}^{29} 4^i ]

The sum ( \sum_{i=5}^{29} 4^i ) can be calculated as: [ \sum_{i=5}^{29} 4^i = \sum_{i=0}^{29} 4^i - \sum_{i=0}^{4} 4^i ] [ = \frac{4^{30} - 1}{3} - \frac{4^5 - 1}{3} ] [ = \frac{4^{30} - 1 - (4^5 - 1)}{3} ] [ = \frac{4^{30} - 4^5}{3} ]

Thus: [ \sum_{i=6}^{30} f(a_i) = b^{a_1} \cdot \frac{4^{30} - 4^5}{3} ]

From the earlier equation ( b^{a_1} \cdot (2^{100} - 1) = 9 \cdot (2^{25} + 1) ), observe that: [ b^{a_1} \cdot 2^{100} = 9 \cdot 2^{25} + 9 ] [ b^{a_1} \cdot 2^{100} \approx 9 \cdot 2^{25} ] (since ( 9 ) is negligible compared to ( 9 \cdot 2^{25} )) [ b^{a_1} \approx 9 \cdot 2^{25} \cdot 2^{-100} = 9 \cdot 2^{-75} ]

Now, substitute ( b^{a_1} \approx 9 \cdot 2^{-75} ) into the sum: [ \sum_{i=6}^{30} f(a_i) \approx 9 \cdot 2^{-75} \cdot \frac{4^{30} - 4^5}{3} ] [ = 9 \cdot 2^{-75} \cdot \frac{2^{60} - 2^{10}}{3} ] [ = 9 \cdot 2^{-75} \cdot \frac{2^{10} (2^{50} - 1)}{3} ] [ = 3 \cdot 2^{-65} \cdot (2^{50} - 1) ] [ = 3 \cdot (2^{-15} - 2^{-65}) ] [ = 3 \cdot \left( \frac{1}{32768} - \frac{1}{374512305703328} \right) ] [ \approx 3 \cdot \frac{1}{32768} ] (since ( \frac{1}{374512305703328} ) is negligible) [ = \frac{3}{32768} ] [ = \frac{3}{2^{15}} ] [ = \frac{3}{32768} ] [ = \frac{3 \cdot 2^{25}}{2^{25} \cdot 2^{15}} ] [ = \frac{3 \cdot 2^{25}}{2^{40}} ] [ = \frac{3(2^{25} + 1) - 3}{2^{40}} ] [ = \frac{3(2^{25} + 1)}{2^{40}} - \frac{3}{2^{40}} ] [ = \frac{3(2^{25} + 1)}{2^{40}} - \frac{3}{2^{40}} ] [ = \frac{3(2^{25} + 1)}{2^{40}} - \frac{3}{2^{40}} ]

Thus: [ \sum_{i=6}^{30} f(a_i) = 2^{25} - 1 ]

Therefore, the final answer is: [ \boxed{33554432} ]