If functions `f:{1,2,…,n} rarr {1995,1996}` satisfying f(1)+f(2)+…+f(1996)=odd integer are formed, the number of such functions can be

Options

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

Answer

To solve the problem, we need to determine the number of functions \( f: \{1, 2, \ldots, n\} \to \{1995, 1996\} \) such that the sum \( f(1) + f(2) + \ldots + f(n) \) is an odd integer.

### Step-by-step Solution:

1. **Understanding the Function Values**:
The function \( f \) can take values either 1995 or 1996 for each input from the set \( \{1, 2, \ldots, n\} \).

2. **Sum of Function Values**:
The sum \( f(1) + f(2) + \ldots + f(n) \) will depend on how many times each value (1995 and 1996) is selected.

3. **Analyzing the Parity**:
- The number 1995 is odd.
- The number 1996 is even.
- The sum of an odd number of odd integers is odd, while the sum of an even number of odd integers is even.

4. **Counting Odd and Even Selections**:
Let \( k \) be the number of times we choose 1995 (odd) and \( n-k \) be the number of times we choose 1996 (even). The total sum can be expressed as:
\[ \text{Sum} = 1995k + 1996(n-k) = 1996n - k \]
For this sum to be odd, \( k \) must be odd (since \( 1996n \) is even).

5. **Possible Values of \( k \)**:
Since \( k \) needs to be odd, the possible values of \( k \) can be \( 1, 3, 5, \ldots, n \) (if \( n \) is odd) or \( 1, 3, 5, \ldots, n-1 \) (if \( n \) is even).

6. **Counting Valid Functions**:
The number of ways to choose \( k \) odd selections from \( n \) is given by \( \binom{n}{k} \). Since \( k \) can take any odd value, we need to sum these combinations:
\[ \text{Total functions} = \sum_{k \text{ odd}} \binom{n}{k} \]

7. **Using Binomial Theorem**:
By the binomial theorem, we know:
\[ (1 + 1)^n = 2^n \quad \text{(sum of all combinations)} \]
and
\[ (1 - 1)^n = 0 \quad \text{(sum of combinations with odd counts)} \]
Thus, the sum of combinations with odd \( k \) is:
\[ \sum_{k \text{ odd}} \binom{n}{k} = \frac{1}{2} \cdot 2^n = 2^{n-1} \]

8. **Final Count of Functions**:
Therefore, the total number of functions \( f \) such that \( f(1) + f(2) + \ldots + f(n) \) is odd is:
\[ 2^{n-1} \]

### Conclusion:
The number of such functions is \( 2^{n-1} \).