If n is a positive integer then `2^(n)gt1+nsqrt((2^(n-1)))`. True or False.

Answer

To determine whether the statement \(2^n > 1 + n\sqrt{2^{n-1}}\) is true or false for positive integers \(n\), we will analyze and simplify the inequality step by step.

### Step 1: Rewrite the Inequality
We start with the inequality:
\[ 2^n > 1 + n\sqrt{2^{n-1}} \]
We can rewrite \(\sqrt{2^{n-1}}\) as \(2^{(n-1)/2}\). Thus, the inequality becomes:
\[ 2^n > 1 + n \cdot 2^{(n-1)/2} \]

### Step 2: Isolate Terms
Next, we can isolate \(2^n\) on one side:
\[ 2^n - n \cdot 2^{(n-1)/2} > 1 \]

### Step 3: Factor Out Common Terms
Notice that \(2^n\) can be expressed as \(2^{(n-1)/2} \cdot 2^{(n+1)/2}\). Therefore, we can factor out \(2^{(n-1)/2}\):
\[ 2^{(n-1)/2}(2^{(n+1)/2} - n) > 1 \]

### Step 4: Analyze the Factor
Now we need to analyze the term \(2^{(n+1)/2} - n\). For \(n = 1\):
\[ 2^{(1+1)/2} - 1 = 2^{1} - 1 = 1 > 0 \]
For \(n = 2\):
\[ 2^{(2+1)/2} - 2 = 2^{3/2} - 2 \approx 2.828 - 2 = 0.828 > 0 \]
For \(n = 3\):
\[ 2^{(3+1)/2} - 3 = 2^{2} - 3 = 4 - 3 = 1 > 0 \]
For \(n = 4\):
\[ 2^{(4+1)/2} - 4 = 2^{5/2} - 4 = 4\sqrt{2} - 4 \approx 5.656 - 4 = 1.656 > 0 \]
As we can see, for small values of \(n\), \(2^{(n+1)/2} - n\) remains positive.

### Step 5: General Case for Larger \(n\)
As \(n\) increases, \(2^{(n+1)/2}\) grows exponentially while \(n\) grows linearly. Thus, for sufficiently large \(n\), \(2^{(n+1)/2} - n > 0\) will hold true.

### Conclusion
Since we have shown that \(2^{(n+1)/2} - n > 0\) for small values of \(n\) and it continues to hold true as \(n\) increases, we conclude that:
\[ 2^n > 1 + n\sqrt{2^{n-1}} \]
is true for all positive integers \(n\).

### Final Answer
**True**