Question

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 __________.

Hint:

For functional equations of the form \( f(x+y) = f(x)f(y) \), assume exponential functions and use series properties if the domain is over reals.

Answer 1

Correct Answer: 96

Step 1: Understand the properties of the function \( f \)

- \( f: \mathbb{R} \to \mathbb{R} \) satisfies:

\qquad - \( f(x)>0 \) for all \( x \in \mathbb{R} \),

\qquad - \( f(x + y) = f(x) f(y) \) for all \( x, y \in \mathbb{R} \).

- The functional equation \( f(x + y) = f(x) f(y) \) suggests \( f \) behaves like an exponential function. Let’s explore this property:

\qquad - Set \( y = 0 \): \( f(x + 0) = f(x) f(0) \implies f(x) = f(x) f(0) \implies f(x) (f(0) - 1) = 0 \).

\qquad - Since \( f(x)>0 \), we have \( f(0) - 1 = 0 \implies f(0) = 1 \).

- Now, consider the functional equation again. Functions satisfying \( f(x + y) = f(x) f(y) \) with \( f(0) = 1 \) and \( f(x)>0 \) are often of the form \( f(x) = e^{kx} \). Let’s assume: \[ f(x) = e^{kx}, \] where \( k \) is a constant. Check:

\qquad - \( f(x + y) = e^{k(x+y)} = e^{kx} e^{ky} = f(x) f(y) \), which satisfies the equation.

\qquad - \( f(x) = e^{kx}>0 \) for all \( x \), which holds for any real \( k \).

- So, \( f(x) = e^{kx} \) is a candidate. We’ll determine \( k \) using the given conditions. Step 2: Use the given condition \( f(a_{31}) = 64 f(a_{25}) \)

- The sequence \( a_1, a_2, \ldots, a_{50} \) is an arithmetic progression (AP).

- Let the first term be \( a_1 = a \), and the common difference be \( d \). Then:

\qquad - \( a_n = a_1 + (n-1)d = a + (n-1)d \),

\qquad - \( a_{25} = a + 24d \),

\qquad - \( a_{31} = a + 30d \).

- Given \( f(a_{31}) = 64 f(a_{25}) \): \[ f(a + 30d) = 64 f(a + 24d). \]
- Substitute \( f(x) = e^{kx} \): \[ e^{k(a + 30d)} = 64 e^{k(a + 24d)}. \]
- Simplify: \[ e^{k(a + 30d) - k(a + 24d)} = 64 \implies e^{k(30d - 24d)} = 64 \implies e^{k \cdot 6d} = 64. \]
- Since \( 64 = 2^6 \), take the natural logarithm: \[ k \cdot 6d = \ln(64) = \ln(2^6) = 6 \ln 2 \implies k d = \ln 2 \implies k = \frac{\ln 2}{d}. \]
- Thus: \[ f(x) = e^{k x} = e^{\left(\frac{\ln 2}{d}\right) x} = \left( e^{\frac{\ln 2}{d}} \right)^x = 2^{x/d}, \] where we used \( e^{\ln 2} = 2 \). So: \[ f(x) = 2^{x/d}. \]
Step 3: Compute the first sum \(\sum_{i=1}^{50} f(a_i)\)

- \( a_i = a + (i-1)d \),

- \( f(a_i) = 2^{a_i / d} = 2^{(a + (i-1)d)/d} = 2^{a/d + (i-1)} = 2^{a/d} \cdot 2^{i-1} \).

- The sum: \[ \sum_{i=1}^{50} f(a_i) = \sum_{i=1}^{50} 2^{a/d} \cdot 2^{i-1} = 2^{a/d} \sum_{i=1}^{50} 2^{i-1}. \]
- The series \(\sum_{i=1}^{50} 2^{i-1}\) is geometric:

\qquad - First term (\( i = 1 \)): \( 2^{1-1} = 2^0 = 1 \),

\qquad - Common ratio: 2,

\qquad - Number of terms: 50.

- Sum of a geometric series \(\sum_{i=1}^{n} r^{i-1} = \frac{r^n - 1}{r - 1}\): \[ \sum_{i=1}^{50} 2^{i-1} = \frac{2^{50} - 1}{2 - 1} = 2^{50} - 1. \]
- So: \[ \sum_{i=1}^{50} f(a_i) = 2^{a/d} (2^{50} - 1). \]
- Given: \[ \sum_{i=1}^{50} f(a_i) = 3 (2^{25} + 1). \]
- Equate: \[ 2^{a/d} (2^{50} - 1) = 3 (2^{25} + 1). \]
- Compute the right-hand side: \[ 2^{25} + 1 \text{ (numerically, } 2^{25} = 33554432, \text{ so } 2^{25} + 1 = 33554433\text{)}, \] \[ 3 (2^{25} + 1) = 3 \times 33554433 = 100663299. \]
- Left-hand side: \[ 2^{50} - 1 = 1125899906842624 - 1 = 1125899906842623. \]
- So: \[ 2^{a/d} (1125899906842623) = 100663299. \]
- Solve for \( 2^{a/d} \): \[ 2^{a/d} = \frac{100663299}{1125899906842623}. \]
- This fraction is small, but we’ll use it later. First, let’s compute the sum we need. Step 4: Compute the sum \(\sum_{i=6}^{30} f(a_i)\)

- Indices from 6 to 30: number of terms = \( 30 - 6 + 1 = 25 \).

- \( f(a_i) = 2^{a/d} \cdot 2^{i-1} \),

- Sum: \[ \sum_{i=6}^{30} f(a_i) = \sum_{i=6}^{30} 2^{a/d} \cdot 2^{i-1} = 2^{a/d} \sum_{i=6}^{30} 2^{i-1}. \]
- Geometric series from \( i = 6 \) to \( 30 \):

\qquad - First term (\( i = 6 \)): \( 2^{6-1} = 2^5 \),

\qquad - Last term (\( i = 30 \)): \( 2^{30-1} = 2^{29} \),

\qquad - Number of terms: 25.

- Sum: \[ \sum_{i=6}^{30} 2^{i-1} = 2^5 + 2^6 + \cdots + 2^{29} = (2^0 + 2^1 + \cdots + 2^{29}) - (2^0 + 2^1 + \cdots + 2^4). \]
- Total from \( 2^0 \) to \( 2^{29} \): \[ \frac{2^{30} - 1}{2 - 1} = 2^{30} - 1. \]
- First five terms (\( 2^0 \) to \( 2^4 \)): \[ \frac{2^5 - 1}{2 - 1} = 2^5 - 1. \]
- So: \[ \sum_{i=6}^{30} 2^{i-1} = (2^{30} - 1) - (2^5 - 1) = 2^{30} - 2^5 = 2^{30} - 32. \]
- Compute: \[ 2^{30} = 1073741824, \quad 2^{30} - 32 = 1073741792. \]
- Thus: \[ \sum_{i=6}^{30} f(a_i) = 2^{a/d} (2^{30} - 32). \]
Step 5: Relate the two sums

- From the given: \[ 2^{a/d} (2^{50} - 1) = 3 (2^{25} + 1). \]
- We need: \[ 2^{a/d} (2^{30} - 32). \]
- Divide the two: \[ \frac{\sum_{i=6}^{30} f(a_i)}{\sum_{i=1}^{50} f(a_i)} = \frac{2^{a/d} (2^{30} - 32)}{2^{a/d} (2^{50} - 1)} = \frac{2^{30} - 32}{2^{50} - 1}. \]
- So: \[ \sum_{i=6}^{30} f(a_i) = \left( \frac{2^{30} - 32}{2^{50} - 1} \right) \cdot 3 (2^{25} + 1). \]
- Numerator ratio: \[ \frac{2^{30} - 32}{2^{50} - 1} = \frac{1073741792}{1125899906842623}. \]
- This fraction is approximately: \[ \frac{1073741792}{1125899906842623} \approx 9.536 \times 10^{-7}. \]
- We already have \( 3 (2^{25} + 1) = 100663299 \).

- Product: \[ 100663299 \times \frac{1073741792}{1125899906842623}. \]
- Notice the pattern in the exponents: \[ 2^{30} - 32 = 2^{30} - 2^5 = 2^5 (2^{25} - 1), \] \[ 2^{50} - 1 = (2^{25} - 1)(2^{25} + 1), \] \[ \frac{2^{30} - 32}{2^{50} - 1} = \frac{2^5 (2^{25} - 1)}{(2^{25} - 1)(2^{25} + 1)} = \frac{2^5}{2^{25} + 1} = \frac{32}{2^{25} + 1}. \]
- So: \[ \sum_{i=6}^{30} f(a_i) = 3 (2^{25} + 1) \cdot \frac{32}{2^{25} + 1} = 3 \cdot 32 = 96. \]
Final Answer: The value of \(\sum_{i=6}^{30} f(a_i)\) is \( \boxed{96} \).