If `(1)/(log_(2)a)+(1)/(log_(4)a)+(1)/(log_(8)a)+(1)/(log_(16)a)+….+`
`(1)/(log_(2^(n))a) = (n(n+1))/(k)` then k `log_(a)2` is equal to

Options

A. 2
B. 3
C. 1
D. 4

Explanation

To solve the problem, we need to evaluate the expression given and find the value of \( k \log_a 2 \).

### Step-by-step Solution:

1. **Rewrite the logarithmic terms**:
We start with the expression:
\[ \frac{1}{\log_2 a} + \frac{1}{\log_4 a} + \frac{1}{\log_8 a} + \frac{1}{\log_{16} a} + \ldots + \frac{1}{\log_{2^n} a} \]
Using the change of base formula for logarithms, we can rewrite each term:
\[ \frac{1}{\log_{2^k} a} = \log_a (2^k) = k \log_a 2 \]
Therefore, we can express the entire sum as:
\[ \sum_{k=1}^{n} k \log_a 2 \]

2. **Factor out the common term**:
The sum can be factored:
\[ \log_a 2 \sum_{k=1}^{n} k \]

3. **Use the formula for the sum of the first n natural numbers**:
The sum of the first \( n \) natural numbers is given by:
\[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \]
Substituting this into our expression gives:
\[ \log_a 2 \cdot \frac{n(n+1)}{2} \]

4. **Set the expression equal to the given equation**:
According to the problem, we have:
\[ \log_a 2 \cdot \frac{n(n+1)}{2} = \frac{n(n+1)}{k} \]

5. **Cross-multiply to solve for k**:
Cross-multiplying gives:
\[ k \cdot \log_a 2 \cdot \frac{n(n+1)}{2} = n(n+1) \]
Simplifying this, we find:
\[ k \log_a 2 = \frac{2n(n+1)}{n(n+1)} = 2 \]

6. **Final result**:
Therefore, we conclude that:
\[ k \log_a 2 = 2 \]

### Conclusion:
The value of \( k \log_a 2 \) is \( 2 \).