Coefficient of `x^5` in `(1+x +x^2 +x^3)^10` is

Options

A. 1910
B. 1902
C. 1819
D. 1932

Answer

To find the coefficient of \( x^5 \) in the expression \( (1 + x + x^2 + x^3)^{10} \), we can follow these steps:

### Step 1: Rewrite the expression
The expression \( 1 + x + x^2 + x^3 \) can be rewritten using the formula for the sum of a geometric series. We can factor it as follows:

\[ 1 + x + x^2 + x^3 = \frac{1 - x^4}{1 - x} \]

Thus, we can express our original expression as:

\[ (1 + x + x^2 + x^3)^{10} = \left(\frac{1 - x^4}{1 - x}\right)^{10} = (1 - x^4)^{10} \cdot (1 - x)^{-10} \]

### Step 2: Expand using the Binomial Theorem
We will use the Binomial Theorem to expand both parts of the expression.

1. **Expanding \( (1 - x^4)^{10} \)**:
\[ (1 - x^4)^{10} = \sum_{k=0}^{10} \binom{10}{k} (-1)^k (x^4)^k = \sum_{k=0}^{10} \binom{10}{k} (-1)^k x^{4k} \]

2. **Expanding \( (1 - x)^{-10} \)**:
\[ (1 - x)^{-10} = \sum_{m=0}^{\infty} \binom{m + 9}{9} x^m \]

### Step 3: Find the coefficient of \( x^5 \)
To find the coefficient of \( x^5 \) in the product of these two expansions, we look for pairs \( (k, m) \) such that:

\[ 4k + m = 5 \]

This gives us the following pairs:
- \( k = 0 \), \( m = 5 \)
- \( k = 1 \), \( m = 1 \)

Now we calculate the contributions from each pair:

1. **For \( k = 0 \), \( m = 5 \)**:
\[ \text{Coefficient} = \binom{10}{0} (-1)^0 \cdot \binom{5 + 9}{9} = 1 \cdot \binom{14}{9} = 2002 \]

2. **For \( k = 1 \), \( m = 1 \)**:
\[ \text{Coefficient} = \binom{10}{1} (-1)^1 \cdot \binom{1 + 9}{9} = 10 \cdot (-1) \cdot \binom{10}{9} = -10 \]

### Step 4: Combine the contributions
Now we sum the contributions from both pairs:

\[ \text{Total Coefficient} = 2002 - 10 = 1992 \]

### Final Answer
Thus, the coefficient of \( x^5 \) in \( (1 + x + x^2 + x^3)^{10} \) is:

\[ \boxed{1992} \]