When `x ^(100)` is divided by `x ^(2) -3x +2,` the remainder is `(2 ^(k +1) -1) x -2 (2 ^(k)-1),` then `k =`

Answer

To solve the problem, we need to find the value of \( k \) when \( x^{100} \) is divided by \( x^2 - 3x + 2 \) and the remainder is given as \( (2^{k+1} - 1)x - 2(2^k - 1) \).

### Step-by-Step Solution:

1. **Identify the Divisor and Remainder:**
- The divisor is \( x^2 - 3x + 2 \).
- The remainder is given as \( (2^{k+1} - 1)x - 2(2^k - 1) \).

2. **Factor the Divisor:**
- The quadratic \( x^2 - 3x + 2 \) can be factored as \( (x - 1)(x - 2) \).

3. **Use the Remainder Theorem:**
- According to the Remainder Theorem, the remainder when dividing \( f(x) \) by \( (x - r) \) is \( f(r) \).
- We will evaluate \( x^{100} \) at the roots of the divisor, which are \( x = 1 \) and \( x = 2 \).

4. **Evaluate at \( x = 1 \):**
- Substitute \( x = 1 \):
\[ 1^{100} = (2^{k+1} - 1) \cdot 1 - 2(2^k - 1) \]
This simplifies to:
\[ 1 = (2^{k+1} - 1) - 2(2^k - 1) \]
\[ 1 = 2^{k+1} - 1 - 2^{k+1} + 2 \]
\[ 1 = 1 \]
- This equation is satisfied, but it doesn't provide new information.

5. **Evaluate at \( x = 2 \):**
- Substitute \( x = 2 \):
\[ 2^{100} = (2^{k+1} - 1) \cdot 2 - 2(2^k - 1) \]
This simplifies to:
\[ 2^{100} = 2(2^{k+1} - 1) - 2(2^k - 1) \]
\[ 2^{100} = 2^{k+2} - 2 - 2^{k+1} + 2 \]
\[ 2^{100} = 2^{k+2} - 2^{k+1} \]
\[ 2^{100} = 2^{k+1}(2 - 1) \]
\[ 2^{100} = 2^{k+1} \]

6. **Equate the Exponents:**
- Since the bases are the same, we can equate the exponents:
\[ 100 = k + 1 \]
- Solving for \( k \):
\[ k = 100 - 1 = 99 \]

### Final Answer:
Thus, the value of \( k \) is \( \boxed{99} \).