`sum_(r=1)^(n) 1/(log_(2^(r))4)` is equal to

Options

A. `(n(n+1))/2`
B. `(n(n+1)(2n+1))/12`
C. `1/(n(n+1))`
D. `(n(n+1))/4`

Explanation

To solve the problem, we need to evaluate the summation:

\[ \sum_{r=1}^{n} \frac{1}{\log_{2^r} 4} \]

### Step 1: Rewrite the logarithm
Using the change of base formula for logarithms, we can rewrite \(\log_{2^r} 4\) as follows:

\[ \log_{2^r} 4 = \frac{\log_2 4}{\log_2 (2^r)} = \frac{\log_2 4}{r} \]

Since \(\log_2 4 = 2\) (because \(2^2 = 4\)), we have:

\[ \log_{2^r} 4 = \frac{2}{r} \]

### Step 2: Substitute back into the summation
Now we can substitute this back into our summation:

\[ \sum_{r=1}^{n} \frac{1}{\log_{2^r} 4} = \sum_{r=1}^{n} \frac{1}{\frac{2}{r}} = \sum_{r=1}^{n} \frac{r}{2} \]

### Step 3: Factor out the constant
We can factor out the constant \(\frac{1}{2}\) from the summation:

\[ \sum_{r=1}^{n} \frac{r}{2} = \frac{1}{2} \sum_{r=1}^{n} r \]

### 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 the formula:

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

### Step 5: Substitute this result back
Substituting this result back into our expression, we get:

\[ \frac{1}{2} \cdot \frac{n(n+1)}{2} = \frac{n(n+1)}{4} \]

### Final Result
Thus, the value of the summation is:

\[ \sum_{r=1}^{n} \frac{1}{\log_{2^r} 4} = \frac{n(n+1)}{4} \]

This matches with option D.