Let S be the sum of the last 24 coefficients in the expansion of `(1 + x)^(47)` when expanded in ascending powers of x, then `(S-2^(44))/(2^(44))`=____

Answer

To solve the problem, we need to find the value of \( S \) which is the sum of the last 24 coefficients in the expansion of \( (1 + x)^{47} \). We will then compute \( \frac{S - 2^{44}}{2^{44}} \).

### Step-by-Step Solution:

1. **Understand the Expansion**:
The expansion of \( (1 + x)^{47} \) can be expressed using the binomial theorem:
\[ (1 + x)^{47} = \sum_{k=0}^{47} \binom{47}{k} x^k \]
where \( \binom{47}{k} \) are the binomial coefficients.

2. **Identify the Last 24 Coefficients**:
The last 24 coefficients correspond to \( k = 24, 25, \ldots, 47 \). Therefore, we can express \( S \) as:
\[ S = \binom{47}{24} + \binom{47}{25} + \cdots + \binom{47}{47} \]

3. **Use the Symmetry of Binomial Coefficients**:
We know that \( \binom{n}{r} = \binom{n}{n-r} \). Thus, we can rewrite the coefficients:
\[ S = \binom{47}{24} + \binom{47}{25} + \cdots + \binom{47}{47} = \binom{47}{23} + \binom{47}{22} + \cdots + \binom{47}{0} \]

4. **Combine the Two Expressions**:
Now, we have:
\[ 2S = \left( \binom{47}{0} + \binom{47}{1} + \cdots + \binom{47}{47} \right) \]
The sum of all binomial coefficients is given by:
\[ \sum_{k=0}^{47} \binom{47}{k} = 2^{47} \]
Therefore:
\[ 2S = 2^{47} \]
This implies:
\[ S = 2^{46} \]

5. **Calculate \( S - 2^{44} \)**:
Now we can compute \( S - 2^{44} \):
\[ S - 2^{44} = 2^{46} - 2^{44} = 2^{44}(2^2 - 1) = 2^{44} \cdot 3 \]

6. **Final Calculation**:
We need to find:
\[ \frac{S - 2^{44}}{2^{44}} = \frac{2^{44} \cdot 3}{2^{44}} = 3 \]

Thus, the final answer is:
\[ \boxed{3} \]