The coefficient of `x^24` in the expansion of `(1+x^2)^(12) (1+x^12)(1+x^24)` is

Options

A. `12C_6`
B. `12C_6+2`
C. `12C_6 + 4`
D. `12C_6 + 6`

Answer

To find the coefficient of \( x^{24} \) in the expansion of \( (1+x^2)^{12} (1+x^{12})(1+x^{24}) \), we can follow these steps:

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

Using the Binomial Theorem, we can expand \( (1+x^2)^{12} \):

\[ (1+x^2)^{12} = \sum_{k=0}^{12} \binom{12}{k} (x^2)^k = \sum_{k=0}^{12} \binom{12}{k} x^{2k} \]

### Step 2: Identify the terms contributing to \( x^{24} \)

Next, we need to consider the other factors \( (1+x^{12})(1+x^{24}) \). The expansion of these factors gives us:

\[ (1+x^{12})(1+x^{24}) = 1 + x^{12} + x^{24} + x^{36} \]

### Step 3: Combine the expansions

Now we need to find the combinations of terms from \( (1+x^2)^{12} \) and \( (1+x^{12})(1+x^{24}) \) that will yield \( x^{24} \).

1. **From \( (1+x^2)^{12} \)**, we can take:
- \( x^{24} \) (which corresponds to \( k=12 \) in \( (1+x^2)^{12} \))
- \( x^{12} \) (which corresponds to \( k=6 \) in \( (1+x^2)^{12} \)) combined with \( x^{12} \) from \( (1+x^{12}) \)
- \( x^0 \) (which corresponds to \( k=0 \) in \( (1+x^2)^{12} \)) combined with \( x^{24} \) from \( (1+x^{24}) \)

### Step 4: Calculate the coefficients

1. The coefficient of \( x^{24} \) from \( (1+x^2)^{12} \) when \( k=12 \) is \( \binom{12}{12} = 1 \).
2. The coefficient of \( x^{12} \) from \( (1+x^2)^{12} \) when \( k=6 \) is \( \binom{12}{6} \).
3. The coefficient of \( x^0 \) from \( (1+x^2)^{12} \) when \( k=0 \) is \( \binom{12}{0} = 1 \).

Combining these contributions, we get:

\[ \text{Coefficient of } x^{24} = 1 + \binom{12}{6} + 1 \]

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

Now, we calculate \( \binom{12}{6} \):

\[ \binom{12}{6} = \frac{12!}{6!6!} = 924 \]

### Step 6: Final calculation

Thus, the coefficient of \( x^{24} \) is:

\[ 1 + 924 + 1 = 926 \]

### Conclusion

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

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