Coefficient of `t^(24)` in `(1+t^(2))^(12)(1+t^(12))(1+t^(24))` is :

Options

A. `""^(12)C_(6)+3`
B. `""^(12)C_(6)+1`
C. `""^(12)C_(6)`
D. `""^(12)C_(6)+2`

Answer

To find the coefficient of \( t^{24} \) in the expression \( (1 + t^2)^{12} (1 + t^{12})(1 + t^{24}) \), we will follow these steps:

### Step 1: Expand \( (1 + t^2)^{12} \)
Using the Binomial Theorem, we can expand \( (1 + t^2)^{12} \):
\[ (1 + t^2)^{12} = \sum_{k=0}^{12} \binom{12}{k} (t^2)^k = \sum_{k=0}^{12} \binom{12}{k} t^{2k} \]
This gives us terms of the form \( t^{2k} \) where \( k = 0, 1, 2, \ldots, 12 \).

### Step 2: Identify the relevant terms
We need to find combinations of terms from \( (1 + t^2)^{12} \), \( (1 + t^{12}) \), and \( (1 + t^{24}) \) that will yield \( t^{24} \).

The possible contributions to \( t^{24} \) can come from:
- \( t^{24} \) from \( (1 + t^{24}) \)
- \( t^{12} \) from \( (1 + t^{12}) \)
- \( t^0 \) from \( (1 + t^2)^{12} \)

### Step 3: Calculate contributions
1. **From \( t^{24} \) in \( (1 + t^{24}) \)**:
- The coefficient of \( t^{24} \) in \( (1 + t^2)^{12} \) corresponds to \( k = 12 \) (since \( 2k = 24 \)):
\[ \text{Coefficient} = \binom{12}{12} = 1 \]

2. **From \( t^{12} \) in \( (1 + t^{12}) \)**:
- The coefficient of \( t^{12} \) in \( (1 + t^2)^{12} \) corresponds to \( k = 6 \) (since \( 2k = 12 \)):
\[ \text{Coefficient} = \binom{12}{6} \]

3. **From \( t^0 \) in \( (1 + t^{12}) \)**:
- The coefficient of \( t^0 \) in \( (1 + t^2)^{12} \) corresponds to \( k = 0 \):
\[ \text{Coefficient} = \binom{12}{0} = 1 \]

### Step 4: Combine contributions
Now, we can combine these contributions:
- From \( t^{24} \): \( 1 \)
- From \( t^{12} \): \( \binom{12}{6} \)
- From \( t^0 \): \( 1 \)

Thus, the total coefficient of \( t^{24} \) is:
\[ 1 + \binom{12}{6} \cdot 1 = 1 + \binom{12}{6} \]

### Step 5: Calculate \( \binom{12}{6} \)
\[ \binom{12}{6} = \frac{12!}{6!6!} = 924 \]

### Final Answer
Therefore, the coefficient of \( t^{24} \) is:
\[ 1 + 924 = 925 \]