`1/(log_(2)a) + 1/(log_(4)a) + 1/(log_(8)a)`+….. To n terms `=(n(n+1))/k`, then k is equal to:

Options

A. `log_(2)a^(2)`
B. `log_(2)(a/2)`
C. `log_(a)2`
D. none of these

Answer

To solve the problem \( \frac{1}{\log_{2} a} + \frac{1}{\log_{4} a} + \frac{1}{\log_{8} a} + \ldots \) up to \( n \) terms, we can follow these steps:

### Step 1: Rewrite the logarithms
We can use the change of base formula for logarithms, which states that \( \log_{b} a = \frac{1}{\log_{a} b} \). Therefore, we can rewrite each term:

\[ \frac{1}{\log_{2} a} = \log_{a} 2, \quad \frac{1}{\log_{4} a} = \log_{a} 4, \quad \frac{1}{\log_{8} a} = \log_{a} 8 \]

Thus, the expression becomes:

\[ \log_{a} 2 + \log_{a} 4 + \log_{a} 8 + \ldots \]

### Step 2: Express the logarithms in terms of powers of 2
We can express \( 4 \) and \( 8 \) as powers of \( 2 \):

\[ \log_{a} 4 = \log_{a} (2^2) = 2 \log_{a} 2, \quad \log_{a} 8 = \log_{a} (2^3) = 3 \log_{a} 2 \]

Continuing this way, we can express the \( k \)-th term as:

\[ \log_{a} (2^k) = k \log_{a} 2 \]

### Step 3: Sum the series
The series now becomes:

\[ \log_{a} 2 \left( 1 + 2 + 3 + \ldots + n \right) \]

The sum \( 1 + 2 + 3 + \ldots + n \) can be calculated using the formula for the sum of the first \( n \) natural numbers:

\[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \]

Thus, we have:

\[ \log_{a} 2 \cdot \frac{n(n+1)}{2} \]

### Step 4: Final expression
Now, substituting this back, we get:

\[ \frac{n(n+1)}{2} \log_{a} 2 \]

### Step 5: Relate to the given equation
According to the problem, this expression is equal to \( \frac{n(n+1)}{k} \). Therefore, we can set:

\[ \frac{n(n+1)}{2} \log_{a} 2 = \frac{n(n+1)}{k} \]

### Step 6: Solve for \( k \)
By comparing both sides, we can find \( k \):

\[ k = 2 \log_{a} 2 \]

### Step 7: Express \( k \) in terms of logarithm base 2
Using the change of base formula again, we can express \( \log_{a} 2 \) as:

\[ \log_{a} 2 = \frac{1}{\log_{2} a} \]

Thus,

\[ k = 2 \cdot \frac{1}{\log_{2} a} = \frac{2}{\log_{2} a} \]

### Conclusion
So, the value of \( k \) is:

\[ k = \log_{2} a^2 \]

### Final Answer
Thus, \( k \) is equal to \( \log_{2} a^2 \).

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