If `x=log_(2^(2))2+log_(2^(3))2^(2)+log_(2^(4))2^(3)+...+log_(2^(n+1))2^(n),` then

Options

A. `xge((1)/(n+1))^(1//n)`
B. `xgen((1)/(n+1))^(1//n)`
C. `xge((n)/(n+1))^(1//n)`
D. none of these

Answer

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

\[ x = \log_{2^2} 2 + \log_{2^3} 2^2 + \log_{2^4} 2^3 + \ldots + \log_{2^{n+1}} 2^n \]

### Step 1: Apply the change of base formula for logarithms

Using the property of logarithms, we can express each term in the sum as follows:

\[ \log_{a^b} c^d = \frac{d}{b} \log_a c \]

So, we can rewrite each term:

\[ \log_{2^2} 2 = \frac{1}{2} \log_2 2 = \frac{1}{2} \]
\[ \log_{2^3} 2^2 = \frac{2}{3} \log_2 2 = \frac{2}{3} \]
\[ \log_{2^4} 2^3 = \frac{3}{4} \log_2 2 = \frac{3}{4} \]
\[ \vdots \]
\[ \log_{2^{n+1}} 2^n = \frac{n}{n+1} \log_2 2 = \frac{n}{n+1} \]

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

\[ x = \frac{1}{2} + \frac{2}{3} + \frac{3}{4} + \ldots + \frac{n}{n+1} \]

### Step 2: Rewrite the expression for \(x\)

Now we can write \(x\) in a more compact form:

\[ x = \sum_{k=1}^{n} \frac{k}{k+1} \]

### Step 3: Simplify the sum

We can simplify each term \(\frac{k}{k+1}\):

\[ \frac{k}{k+1} = 1 - \frac{1}{k+1} \]

So, we have:

\[ x = \sum_{k=1}^{n} \left( 1 - \frac{1}{k+1} \right) = \sum_{k=1}^{n} 1 - \sum_{k=1}^{n} \frac{1}{k+1} \]

Calculating the first sum:

\[ \sum_{k=1}^{n} 1 = n \]

And the second sum:

\[ \sum_{k=1}^{n} \frac{1}{k+1} = \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n+1} \]

### Step 4: Combine the results

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

\[ x = n - \left( \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n+1} \right) \]

### Step 5: Apply the AM-GM inequality

Now we can apply the AM-GM inequality to the series:

\[ \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \ldots, \frac{n}{n+1} \]

The arithmetic mean (AM) of these terms is:

\[ \text{AM} = \frac{\frac{1}{2} + \frac{2}{3} + \frac{3}{4} + \ldots + \frac{n}{n+1}}{n} \]

And the geometric mean (GM) is:

\[ \text{GM} = \left( \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac{n}{n+1} \right)^{\frac{1}{n}} \]

### Step 6: Conclude the inequality

Using AM-GM, we have:

\[ \text{AM} \geq \text{GM} \]

This leads us to:

\[ \frac{x}{n} \geq \left( \frac{1}{2} \cdot \frac{2}{3} \cdots \frac{n}{n+1} \right)^{\frac{1}{n}} \]

Thus, we can conclude that:

\[ x \geq n \cdot \left( \frac{1}{n+1} \right)^{\frac{1}{n}} \]

### Final Result

Therefore, the final result is:

\[ x \geq \frac{n}{n+1}^{\frac{1}{n}} \]