In a poisson distribution, the variance is m. The sum of terms in odd places in the distribution is

Options

A. `e^(-m)`
B. `e^(-m)cos h (m)`
C. `e^(-m)sin h(m)`
D. `e^(-m)cot h(m)`

Answer

To solve the problem step by step, we need to find the sum of the terms in odd places of a Poisson distribution where the variance is \( m \).

### Step-by-Step Solution:

1. **Understanding the Poisson Distribution**:
The probability mass function of a Poisson distribution is given by:
\[ f(x) = \frac{e^{-\lambda} \lambda^x}{x!} \]
where \( \lambda \) is the mean and variance. Given that the variance is \( m \), we have \( \lambda = m \).

2. **Setting Up the Function**:
Substituting \( \lambda \) with \( m \), the function becomes:
\[ f(x) = \frac{e^{-m} m^x}{x!} \]

3. **Identifying Terms in Odd Places**:
We need to find the sum of the terms at odd places, which corresponds to \( x = 1, 3, 5, \ldots \). Thus, we can express the sum as:
\[ S = f(1) + f(3) + f(5) + \ldots \]

4. **Calculating the Sum**:
The sum can be expressed as:
\[ S = e^{-m} \left( \frac{m^1}{1!} + \frac{m^3}{3!} + \frac{m^5}{5!} + \ldots \right) \]
This series can be recognized as the Taylor series expansion for the hyperbolic sine function, \( \sinh(x) \):
\[ \sinh(m) = \frac{e^m - e^{-m}}{2} \]
Therefore, we can rewrite the sum as:
\[ S = e^{-m} \cdot \sinh(m) \]

5. **Expressing in Terms of Exponential Functions**:
The hyperbolic sine function can also be expressed in terms of exponential functions:
\[ S = e^{-m} \cdot \frac{e^m - e^{-m}}{2} = \frac{1}{2} (1 - e^{-2m}) \]

6. **Final Expression**:
Thus, the sum of the terms in odd places in the Poisson distribution is:
\[ S = e^{-m} \cdot \sinh(m) = e^{-m} \cdot \frac{e^m - e^{-m}}{2} = \frac{1}{2} (1 - e^{-2m}) \]

### Final Answer:
The sum of the terms in odd places in the Poisson distribution is:
\[ S = e^{-m} \cdot \sinh(m) \]