The number of functions `f: {1,2, 3,... n}-> {2016, 2017}`, where ne N, which satisfy thecondition f1)+f(2)+ ...+ f(n) is an odd number are

Options

A. `2^(n)`
B. `n*2^(n-1)`
C. `2^(n-1)`
D. n!

Answer

To find the number of functions \( f: \{1, 2, 3, \ldots, n\} \to \{2016, 2017\} \) such that the sum \( f(1) + f(2) + \ldots + f(n) \) is an odd number, we can follow these steps:

### Step 1: Understand the problem
We need to find functions that map from a set of \( n \) natural numbers to the set \( \{2016, 2017\} \). The sum of the function values must be odd.

### Step 2: Identify the properties of the numbers
- The number 2016 is even.
- The number 2017 is odd.

### Step 3: Analyze the sum condition
For the sum \( f(1) + f(2) + \ldots + f(n) \) to be odd, we need to have an odd number of odd values (2017) in the sum. This means:
- If we have an even number of 2017s, the sum will be even.
- If we have an odd number of 2017s, the sum will be odd.

### Step 4: Count the functions
Let \( k \) be the number of times 2017 appears in the function values. For the sum to be odd, \( k \) must be odd. The possible odd values for \( k \) can be \( 1, 3, 5, \ldots, n \).

### Step 5: Calculate the number of functions for each odd \( k \)
For each odd \( k \):
- Choose \( k \) positions out of \( n \) to assign the value 2017. This can be done in \( \binom{n}{k} \) ways.
- The remaining \( n-k \) positions will automatically be assigned the value 2016.

### Step 6: Sum over all odd \( k \)
We need to sum \( \binom{n}{k} \) for all odd \( k \):
\[ \text{Total functions} = \sum_{k \text{ odd}} \binom{n}{k} \]

### Step 7: Use the binomial theorem
From the binomial theorem, we know:
\[ (1 + 1)^n = \sum_{k=0}^{n} \binom{n}{k} = 2^n \]
\[ (1 - 1)^n = \sum_{k=0}^{n} (-1)^k \binom{n}{k} = 0 \]
Adding these two equations gives:
\[ 2^n = \sum_{k \text{ even}} \binom{n}{k} + \sum_{k \text{ odd}} \binom{n}{k} \]
Subtracting the second equation from the first gives:
\[ 0 = \sum_{k \text{ even}} \binom{n}{k} - \sum_{k \text{ odd}} \binom{n}{k} \]
This implies:
\[ \sum_{k \text{ odd}} \binom{n}{k} = \sum_{k \text{ even}} \binom{n}{k} = \frac{2^n}{2} = 2^{n-1} \]

### Step 8: Conclusion
Thus, the total number of functions \( f \) such that \( f(1) + f(2) + \ldots + f(n) \) is odd is:
\[ \boxed{2^{n-1}} \]