The coefficient of `x^(4)` in the expansion of `(1+x+x^(2)+x^(3))^(n)` is

Options

A. `""^(n)C_(4)`
B. `""^(n)C_(4)+""^(n)C_(2)`
C. `""^(n)C_(4).""^(n)C_(2)`
D. `""^(n)C_(2)+""^(n)C_(2)+""^(n)C_(1).""^(n)C_(2)`

Answer

To find the coefficient of \( x^4 \) in the expansion of \( (1 + x + x^2 + x^3)^n \), we can follow these steps:

### Step 1: Rewrite the expression
We can rewrite \( 1 + x + x^2 + x^3 \) as a geometric series:
\[ 1 + x + x^2 + x^3 = \frac{1 - x^4}{1 - x} \quad \text{(for } |x| < 1\text{)} \]
Thus, we can express the original expression as:
\[ (1 + x + x^2 + x^3)^n = \left(\frac{1 - x^4}{1 - x}\right)^n \]

### Step 2: Expand using the Binomial Theorem
Using the Binomial Theorem, we can expand \( \left(\frac{1 - x^4}{1 - x}\right)^n \):
\[ \left(1 - x^4\right)^n \cdot (1 - x)^{-n} \]
Using the Binomial expansion:
\[ (1 - x^4)^n = \sum_{k=0}^{n} \binom{n}{k} (-1)^k (x^4)^k = \sum_{k=0}^{n} \binom{n}{k} (-1)^k x^{4k} \]
And for \( (1 - x)^{-n} \):
\[ (1 - x)^{-n} = \sum_{m=0}^{\infty} \binom{n + m - 1}{m} x^m \]

### Step 3: Find the coefficient of \( x^4 \)
To find the coefficient of \( x^4 \) in the product of the two expansions, we need to consider the contributions from both series:
- From \( (1 - x^4)^n \), we can take \( k = 0 \) (which contributes \( \binom{n}{0} (-1)^0 x^0 = 1 \)) and we need the coefficient of \( x^4 \) from \( (1 - x)^{-n} \), which is \( \binom{n + 4 - 1}{4} = \binom{n + 3}{4} \).
- From \( (1 - x^4)^n \), we can take \( k = 1 \) (which contributes \( \binom{n}{1} (-1)^1 x^4 \)), and we need the coefficient of \( x^0 \) from \( (1 - x)^{-n} \), which is \( \binom{n + 0 - 1}{0} = 1 \).

Thus, the coefficient of \( x^4 \) is:
\[ \text{Coefficient of } x^4 = \binom{n + 3}{4} - \binom{n}{1} \]

### Final Result
The coefficient of \( x^4 \) in the expansion of \( (1 + x + x^2 + x^3)^n \) is:
\[ \binom{n + 3}{4} - n \]