If `f(n)=prod_(i=2)^(n-1)log_i(i+1)`, the value of `sum_(k=1)^100f(2^k)` equals

Options

A. 5010
B. 5050
C. 5100
D. 5049

Answer

To solve the problem, we need to evaluate the function \( f(n) \) defined as:

\[ f(n) = \prod_{i=2}^{n-1} \log_i(i+1) \]

And then find the value of:

\[ \sum_{k=1}^{100} f(2^k) \]

### Step 1: Express \( f(n) \)

We start with the expression for \( f(n) \):

\[ f(n) = \prod_{i=2}^{n-1} \log_i(i+1) \]

Using the change of base formula for logarithms, we can rewrite \( \log_i(i+1) \) as:

\[ \log_i(i+1) = \frac{\log(i+1)}{\log(i)} \]

Thus, we can express \( f(n) \) as:

\[ f(n) = \prod_{i=2}^{n-1} \frac{\log(i+1)}{\log(i)} \]

### Step 2: Simplify the product

Now, we can expand the product:

\[ f(n) = \frac{\log(3)}{\log(2)} \cdot \frac{\log(4)}{\log(3)} \cdot \frac{\log(5)}{\log(4)} \cdots \frac{\log(n)}{\log(n-1)} \]

Notice that in this product, all intermediate terms cancel out, leaving us with:

\[ f(n) = \frac{\log(n)}{\log(2)} \]

### Step 3: Evaluate \( f(2^k) \)

Now, we substitute \( n = 2^k \):

\[ f(2^k) = \frac{\log(2^k)}{\log(2)} = \frac{k \log(2)}{\log(2)} = k \]

### Step 4: Calculate the summation

Now we need to calculate:

\[ \sum_{k=1}^{100} f(2^k) = \sum_{k=1}^{100} k \]

Using the formula for the sum of the first \( n \) natural numbers:

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

For \( n = 100 \):

\[ \sum_{k=1}^{100} k = \frac{100 \times 101}{2} = 5050 \]

### Final Answer

Thus, the value of \( \sum_{k=1}^{100} f(2^k) \) is:

\[ \boxed{5050} \]