For a random variable X if P(X=k) = `4((k+1)/5^k)a`, for k=0,1,2,…then a= _____

Answer

To find the value of \( a \) for the random variable \( X \) defined by the probability mass function \( P(X=k) = 4\left(\frac{k+1}{5^k}\right)a \) for \( k=0,1,2,\ldots \), we need to ensure that the total probability sums to 1.

### Step-by-Step Solution:

1. **Set Up the Equation for Total Probability:**
The total probability must equal 1:
\[ \sum_{k=0}^{\infty} P(X=k) = 1 \]
Therefore, we can write:
\[ \sum_{k=0}^{\infty} 4\left(\frac{k+1}{5^k}\right)a = 1 \]

2. **Factor Out \( 4a \):**
We can factor out \( 4a \) from the summation:
\[ 4a \sum_{k=0}^{\infty} \frac{k+1}{5^k} = 1 \]

3. **Calculate the Summation:**
We need to evaluate the summation \( \sum_{k=0}^{\infty} \frac{k+1}{5^k} \). This can be split into two parts:
\[ \sum_{k=0}^{\infty} \frac{k+1}{5^k} = \sum_{k=0}^{\infty} \frac{k}{5^k} + \sum_{k=0}^{\infty} \frac{1}{5^k} \]
The second part, \( \sum_{k=0}^{\infty} \frac{1}{5^k} \), is a geometric series:
\[ \sum_{k=0}^{\infty} \frac{1}{5^k} = \frac{1}{1 - \frac{1}{5}} = \frac{5}{4} \]

4. **Calculate \( \sum_{k=0}^{\infty} \frac{k}{5^k} \):**
To find \( \sum_{k=0}^{\infty} \frac{k}{5^k} \), we can use the formula for the sum of a series:
\[ \sum_{k=0}^{\infty} kx^k = \frac{x}{(1-x)^2} \]
where \( x = \frac{1}{5} \):
\[ \sum_{k=0}^{\infty} k\left(\frac{1}{5}\right)^k = \frac{\frac{1}{5}}{\left(1 - \frac{1}{5}\right)^2} = \frac{\frac{1}{5}}{\left(\frac{4}{5}\right)^2} = \frac{1}{5} \cdot \frac{25}{16} = \frac{5}{16} \]

5. **Combine the Results:**
Now we can combine both parts:
\[ \sum_{k=0}^{\infty} \frac{k+1}{5^k} = \frac{5}{16} + \frac{5}{4} = \frac{5}{16} + \frac{20}{16} = \frac{25}{16} \]

6. **Substitute Back into the Equation:**
Now substitute back into the equation:
\[ 4a \cdot \frac{25}{16} = 1 \]

7. **Solve for \( a \):**
Rearranging gives:
\[ a = \frac{16}{100} = \frac{4}{25} \]

### Final Answer:
Thus, the value of \( a \) is:
\[ \boxed{\frac{4}{25}} \]