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)/(lambda))` then `lambda` equals

Options

A. `"log"_(2)a`
B. `"log"_(a)4`
C. `"log"_(2)a^(2)`
D. none of these

Explanation

To solve the equation

\[ \frac{1}{\log_2 a} + \frac{1}{\log_4 a} + \frac{1}{\log_8 a} + \frac{1}{\log_{16} a} + \ldots + \frac{1}{\log_{2^n} a} = \frac{n(n+1)}{\lambda} \]

we will follow these steps:

### Step 1: Rewrite the logarithms
Using the change of base formula, we know that:

\[ \log_b a = \frac{1}{\log_a b} \]

Thus, we can rewrite each term on the left-hand side:

\[ \frac{1}{\log_{2^k} a} = \log_a (2^k) \]

So, we can rewrite the entire left-hand side:

\[ \log_a 2 + \log_a 4 + \log_a 8 + \ldots + \log_a (2^n) \]

### Step 2: Simplify the logarithmic terms
Next, we can express the logarithms in terms of base 2:

\[ \log_a (2^k) = k \cdot \log_a 2 \]

Thus, the left-hand side becomes:

\[ \log_a 2 + 2 \log_a 2 + 3 \log_a 2 + \ldots + n \log_a 2 \]

### Step 3: Factor out the common term
Factoring out \(\log_a 2\), we get:

\[ \log_a 2 \left(1 + 2 + 3 + \ldots + n\right) \]

### Step 4: Use the formula for the sum of the first n natural numbers
The sum of the first n natural numbers is given by:

\[ 1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2} \]

Thus, we can rewrite the left-hand side as:

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

### Step 5: Set the left-hand side equal to the right-hand side
Now we have:

\[ \log_a 2 \cdot \frac{n(n+1)}{2} = \frac{n(n+1)}{\lambda} \]

### Step 6: Cancel out \(n(n+1)\) (assuming \(n(n+1) \neq 0\))
Dividing both sides by \(n(n+1)\):

\[ \frac{\log_a 2}{2} = \frac{1}{\lambda} \]

### Step 7: Solve for \(\lambda\)
Taking the reciprocal gives:

\[ \lambda = \frac{2}{\log_a 2} \]

### Step 8: Rewrite \(\log_a 2\) using change of base formula
Using the change of base formula:

\[ \log_a 2 = \frac{1}{\log_2 a} \]

Thus, we can express \(\lambda\) as:

\[ \lambda = 2 \cdot \log_2 a \]

### Conclusion
Therefore, the value of \(\lambda\) is:

\[ \lambda = 2 \log_2 a \]