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 can break down the problem step by step.

### Step 1: Expand \( (1 + t^2)^{12} \)

Using the binomial theorem, we can expand \( (1 + t^2)^{12} \):
\[ (1 + t^2)^{12} = \sum_{r=0}^{12} \binom{12}{r} (t^2)^r = \sum_{r=0}^{12} \binom{12}{r} t^{2r} \]
This means we have terms of the form \( \binom{12}{r} t^{2r} \) for \( r = 0, 1, 2, \ldots, 12 \).

### Step 2: Consider the other factors \( (1 + t^{12})(1 + t^{24}) \)

Next, we can expand the product \( (1 + t^{12})(1 + t^{24}) \):
\[ (1 + t^{12})(1 + t^{24}) = 1 + t^{12} + t^{24} + t^{36} \]
This means we have four possible contributions to the coefficient of \( t^{24} \): from the constant term (1), from \( t^{12} \), from \( t^{24} \), and from \( t^{36} \).

### Step 3: Combine contributions to find \( t^{24} \)

Now, we will find the contributions to \( t^{24} \) from each term in the expansion of \( (1 + t^2)^{12} \):

1. **From the constant term (1)**:
- We need \( 2r = 24 \) which gives \( r = 12 \).
- Contribution: \( \binom{12}{12} = 1 \).

2. **From \( t^{12} \)**:
- We need \( 2r + 12 = 24 \) which gives \( 2r = 12 \) or \( r = 6 \).
- Contribution: \( \binom{12}{6} \).

3. **From \( t^{24} \)**:
- We need \( 2r + 24 = 24 \) which gives \( 2r = 0 \) or \( r = 0 \).
- Contribution: \( \binom{12}{0} = 1 \).

4. **From \( t^{36} \)**:
- We need \( 2r + 36 = 24 \) which gives \( 2r = -12 \) (not possible, so contribution is 0).

### Step 4: Sum the contributions

Now we sum all the contributions:
\[ \text{Total Coefficient} = \binom{12}{12} + \binom{12}{6} + \binom{12}{0} = 1 + \binom{12}{6} + 1 \]
\[ = 2 + \binom{12}{6} \]

### Step 5: Calculate \( \binom{12}{6} \)

Using the formula for combinations:
\[ \binom{12}{6} = \frac{12!}{6!6!} = 924 \]

### Final Calculation

Thus, the total coefficient of \( t^{24} \) is:
\[ 2 + 924 = 926 \]

### Answer

The coefficient of \( t^{24} \) in \( (1 + t^2)^{12} (1 + t^{12})(1 + t^{24}) \) is **926**.

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