If a,b are distinct rational numbers, then for all `n in N` the number `a^(n)-b^(n)` is divisible by

Options

A. a-b
B. a+b
C. 2a-b
D. a-2b

Answer

To solve the problem, we need to show that for distinct rational numbers \( a \) and \( b \), the expression \( a^n - b^n \) is divisible by \( a - b \) for all natural numbers \( n \).

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression \( a^n - b^n \). We want to show that this expression can be factored in such a way that \( a - b \) is a factor.

2. **Using the Binomial Theorem**:
We can express \( a^n \) in a different form:
\[ a^n = (a - b + b)^n \]
By applying the Binomial Theorem, we can expand this:
\[ (x + y)^n = \sum_{r=0}^{n} \binom{n}{r} x^{n-r} y^r \]
Here, let \( x = a - b \) and \( y = b \).

3. **Expanding the Expression**:
Expanding \( (a - b + b)^n \) gives:
\[ a^n = \sum_{r=0}^{n} \binom{n}{r} (a - b)^{n-r} b^r \]
The first term of this expansion is \( (a - b)^n \) when \( r = 0 \), and the last term is \( b^n \) when \( r = n \).

4. **Finding \( a^n - b^n \)**:
Now, we can write:
\[ a^n - b^n = \sum_{r=0}^{n} \binom{n}{r} (a - b)^{n-r} b^r - b^n \]
This simplifies to:
\[ a^n - b^n = (a - b) \left( \sum_{r=0}^{n-1} \binom{n}{r+1} (a - b)^{n-1-r} b^{r} \right) \]

5. **Factoring Out \( a - b \)**:
From the above expression, we can clearly see that \( a^n - b^n \) can be factored as:
\[ a^n - b^n = (a - b) \cdot Q(n) \]
where \( Q(n) \) is some polynomial in \( a \) and \( b \).

6. **Conclusion**:
Since \( a - b \) is a factor of \( a^n - b^n \), we conclude that \( a^n - b^n \) is divisible by \( a - b \) for all \( n \in \mathbb{N} \).

### Final Answer:
Thus, we have shown that \( a^n - b^n \) is divisible by \( a - b \).

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