Let P(n) be the statement :
`3^(n)gt n`
What is `P(n+1)` ?

Answer

To find \( P(n+1) \) given that \( P(n) \) is the statement \( 3^n > n \), we can follow these steps:

1. **Understand the statement \( P(n) \)**:
The statement \( P(n) \) is defined as \( 3^n > n \). This means that for any integer \( n \), the value of \( 3^n \) is greater than \( n \).

2. **Define \( P(n+1) \)**:
To find \( P(n+1) \), we need to replace \( n \) in the original statement \( P(n) \) with \( n+1 \). Therefore, we have:
\[ P(n+1) : 3^{n+1} > n+1 \]

3. **Rewrite \( P(n+1) \)**:
We can express \( 3^{n+1} \) using the properties of exponents:
\[ 3^{n+1} = 3^n \cdot 3 \]
Thus, we can rewrite \( P(n+1) \) as:
\[ 3^n \cdot 3 > n + 1 \]

4. **Relate \( P(n+1) \) to \( P(n) \)**:
Since we know from \( P(n) \) that \( 3^n > n \), we can use this information. If \( 3^n > n \), then multiplying both sides of this inequality by 3 gives:
\[ 3 \cdot 3^n > 3n \]
Therefore, we have:
\[ 3^{n+1} > 3n \]

5. **Compare \( 3n \) and \( n + 1 \)**:
Now we need to show that \( 3n > n + 1 \) for \( n \geq 1 \):
\[ 3n > n + 1 \implies 3n - n > 1 \implies 2n > 1 \implies n > \frac{1}{2} \]
Since \( n \) is an integer and we are considering \( n \geq 1 \), this inequality holds true.

6. **Conclusion**:
Therefore, we have shown that if \( P(n) \) is true, then \( P(n+1) \) is also true. Thus, we can conclude:
\[ P(n+1) : 3^{n+1} > n + 1 \]