If X is a random variable with probability distribution `P(X=k)=((k+1)C)/(2^(k)),K=0,1,2,….` then find C.

Answer

To find the value of the constant \( C \) in the probability distribution \( P(X=k) = \frac{(k+1)C}{2^k} \) for \( k = 0, 1, 2, \ldots \), we need to ensure that the total probability sums to 1. This is a fundamental property of probability distributions.

### Step-by-Step Solution:

1. **Set Up the Equation for Total Probability:**
\[ \sum_{k=0}^{\infty} P(X=k) = 1 \]
Substituting the expression for \( P(X=k) \):
\[ \sum_{k=0}^{\infty} \frac{(k+1)C}{2^k} = 1 \]

2. **Factor Out the Constant \( C \):**
\[ C \sum_{k=0}^{\infty} \frac{(k+1)}{2^k} = 1 \]

3. **Evaluate the Series \( \sum_{k=0}^{\infty} \frac{(k+1)}{2^k} \):**
We can split this series into two parts:
\[ \sum_{k=0}^{\infty} \frac{(k+1)}{2^k} = \sum_{k=0}^{\infty} \frac{k}{2^k} + \sum_{k=0}^{\infty} \frac{1}{2^k} \]
The second part is a geometric series:
\[ \sum_{k=0}^{\infty} \frac{1}{2^k} = \frac{1}{1 - \frac{1}{2}} = 2 \]

4. **Evaluate \( \sum_{k=0}^{\infty} \frac{k}{2^k} \):**
This can be derived using the formula for the sum of a series:
\[ \sum_{k=0}^{\infty} kx^k = \frac{x}{(1-x)^2} \quad \text{for } |x| < 1 \]
Setting \( x = \frac{1}{2} \):
\[ \sum_{k=0}^{\infty} k \left(\frac{1}{2}\right)^k = \frac{\frac{1}{2}}{(1 - \frac{1}{2})^2} = \frac{\frac{1}{2}}{\left(\frac{1}{2}\right)^2} = 2 \]

5. **Combine the Results:**
Therefore,
\[ \sum_{k=0}^{\infty} \frac{(k+1)}{2^k} = 2 + 2 = 4 \]

6. **Substitute Back into the Equation:**
Now we have:
\[ C \cdot 4 = 1 \]

7. **Solve for \( C \):**
\[ C = \frac{1}{4} \]

### Final Answer:
The value of the constant \( C \) is \( \frac{1}{4} \).