Sum of the last 12 coefficients in the binomial expansion of `(1 + x)^(23)` is:

Options

A. `2^(22)`
B. `2^(23)`
C. `2^(23)-""^(23)C_(11)`
D. `2^(23)-2^(11)`

Answer

To find the sum of the last 12 coefficients in the binomial expansion of \((1 + x)^{23}\), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Binomial Expansion**:
The binomial expansion of \((1 + x)^n\) is given by:
\[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]
For our case, \(n = 23\).

2. **Identify the Coefficients**:
The coefficients in the expansion are \(\binom{23}{0}, \binom{23}{1}, \ldots, \binom{23}{23}\).

3. **Count the Total Number of Terms**:
The total number of terms in the expansion is \(n + 1\), which is \(23 + 1 = 24\).

4. **Identify the Last 12 Coefficients**:
The last 12 coefficients correspond to:
\[ \binom{23}{12}, \binom{23}{13}, \ldots, \binom{23}{23} \]

5. **Sum of the Last 12 Coefficients**:
To find the sum of these coefficients, we can use the property of binomial coefficients:
\[ \sum_{k=r}^{n} \binom{n}{k} = 2^n - \sum_{k=0}^{r-1} \binom{n}{k} \]
Here, \(r = 12\) and \(n = 23\). Thus, we need to calculate:
\[ \sum_{k=12}^{23} \binom{23}{k} = 2^{23} - \sum_{k=0}^{11} \binom{23}{k} \]

6. **Calculate the Total Sum**:
Since \(\sum_{k=0}^{11} \binom{23}{k} = \sum_{k=12}^{23} \binom{23}{k}\) (by symmetry of binomial coefficients), we can say:
\[ \sum_{k=12}^{23} \binom{23}{k} = \frac{1}{2} \cdot 2^{23} = 2^{22} \]

7. **Final Result**:
Therefore, the sum of the last 12 coefficients in the binomial expansion of \((1 + x)^{23}\) is:
\[ 2^{22} \]