If `agtbgt1` then `a^(n)-b^(n)gtn(a-b)` for every +ive integer `n ge2`.True or False.

Answer

To determine whether the statement "If \( a > b > 1 \) then \( a^n - b^n > n(a - b) \) for every positive integer \( n \geq 2 \)" is true or false, we will analyze the inequality step by step.

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
We are given that \( a > b > 1 \). This means both \( a \) and \( b \) are positive numbers greater than 1.

2. **Consider the Expression**:
We need to analyze the expression \( a^n - b^n \) and compare it to \( n(a - b) \).

3. **Using the Binomial Theorem**:
The expression \( a^n - b^n \) can be factored using the difference of powers:
\[ a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \ldots + b^{n-1}) \]
Here, \( a^{n-1} + a^{n-2}b + \ldots + b^{n-1} \) is a sum of \( n \) terms.

4. **Analyzing the Sum**:
Since \( a > b > 1 \), each term in the sum \( a^{n-1}, a^{n-2}b, \ldots, b^{n-1} \) is positive. Therefore, the entire sum \( a^{n-1} + a^{n-2}b + \ldots + b^{n-1} \) is greater than \( n \cdot b^{n-1} \) (since \( b^{n-1} \) is the smallest term in the sum).

5. **Inequality Formation**:
Thus, we can write:
\[ a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + \ldots + b^{n-1}) > (a - b)(n \cdot b^{n-1}) \]

6. **Comparing with \( n(a - b) \)**:
We need to show that:
\[ (a - b)(n \cdot b^{n-1}) > n(a - b) \]
Dividing both sides by \( a - b \) (which is positive since \( a > b \)):
\[ n \cdot b^{n-1} > n \]
This simplifies to:
\[ b^{n-1} > 1 \]
Since \( b > 1 \), raising \( b \) to any positive power \( n-1 \) (where \( n \geq 2 \)) will indeed yield a result greater than 1.

7. **Conclusion**:
Therefore, we conclude that:
\[ a^n - b^n > n(a - b) \]
is true for every positive integer \( n \geq 2 \).

### Final Answer:
The statement is **True**.