Coefficient of `x^(2)`in the expansion of `(x^(3) + 2x^(2) + x + 4)^(15)` is
(a) Prime (b) Composite (c) 0 (d) Perfect square

Options

A. Prime
B. Composite
C. 0
D. Perfect square

Explanation

To find the coefficient of \( x^2 \) in the expansion of \( (x^3 + 2x^2 + x + 4)^{15} \), we can use the multinomial expansion. The general term in the expansion can be expressed as:

\[ \frac{15!}{k_1! k_2! k_3! k_4!} (x^3)^{k_1} (2x^2)^{k_2} (x)^{k_3} (4)^{k_4} \]

where \( k_1 + k_2 + k_3 + k_4 = 15 \).

To find the coefficient of \( x^2 \), we need to consider the powers of \( x \) contributed by each term:

- From \( (x^3)^{k_1} \), we get \( 3k_1 \).
- From \( (2x^2)^{k_2} \), we get \( 2k_2 \).
- From \( (x)^{k_3} \), we get \( k_3 \).
- From \( (4)^{k_4} \), we get \( 0 \) (since it does not contribute to the power of \( x \)).

We need to satisfy the equation:

\[ 3k_1 + 2k_2 + k_3 = 2 \]

Now, since \( k_1 + k_2 + k_3 + k_4 = 15 \), we can express \( k_4 \) as:

\[ k_4 = 15 - k_1 - k_2 - k_3 \]

Next, we will analyze the possible values of \( k_1, k_2, k_3 \):

1. **Case 1**: \( k_1 = 0 \)
- Then \( 2k_2 + k_3 = 2 \).
- Possible pairs \((k_2, k_3)\):
- \( (1, 0) \) → \( k_2 = 1, k_3 = 0 \) → \( k_4 = 14 \)
- \( (0, 2) \) → \( k_2 = 0, k_3 = 2 \) → \( k_4 = 13 \)

2. **Case 2**: \( k_1 = 1 \)
- Then \( 3 + 2k_2 + k_3 = 2 \) → No valid solution since \( k_1 = 1 \) gives \( 3 \) already.

3. **Case 3**: \( k_1 = 2 \)
- Then \( 6 + 2k_2 + k_3 = 2 \) → No valid solution since \( k_1 = 2 \) gives \( 6 \) already.

Now we will calculate the coefficient for the valid pairs found in Case 1:

- For \( (k_2, k_3) = (1, 0) \):
\[ \text{Coefficient} = \frac{15!}{0! \cdot 1! \cdot 0! \cdot 14!} (x^3)^0 (2x^2)^1 (x)^0 (4)^{14} = \frac{15!}{1 \cdot 14!} \cdot 2 \cdot 4^{14} \]
\[ = 15 \cdot 2 \cdot 4^{14} = 30 \cdot 4^{14} \]

- For \( (k_2, k_3) = (0, 2) \):
\[ \text{Coefficient} = \frac{15!}{0! \cdot 0! \cdot 2! \cdot 13!} (x^3)^0 (2x^2)^0 (x)^2 (4)^{13} = \frac{15!}{2! \cdot 13!} \cdot 4^{13} \]
\[ = \frac{15 \cdot 14}{2} \cdot 4^{13} = 105 \cdot 4^{13} \]

Now, we need to combine these coefficients:

\[ \text{Total Coefficient} = 30 \cdot 4^{14} + 105 \cdot 4^{13} = 30 \cdot 4^{13} \cdot 4 + 105 \cdot 4^{13} = (120 + 105) \cdot 4^{13} = 225 \cdot 4^{13} \]

Next, we need to check if \( 225 \cdot 4^{13} \) is a perfect square:

- \( 225 = 15^2 \) (which is a perfect square).
- \( 4^{13} = (2^2)^{13} = 2^{26} \) (which is also a perfect square).

Thus, \( 225 \cdot 4^{13} \) is a perfect square.

Finally, the answer is:

**Answer**: (d) Perfect square.