lf `(1 + x + x^2 + x^3)^5 = a_0+a_1x +a_2x^2+.....+a_(15)x^15`, then `a_(10)` equals to

Options

A. 99
B. 100
C. 101
D. 110

Answer

To find the coefficient \( a_{10} \) in the expansion of \( (1 + x + x^2 + x^3)^5 \), we can start by rewriting the expression in a more manageable form.

### Step-by-Step Solution:

1. **Rewrite the Expression**:
\[ (1 + x + x^2 + x^3)^5 = (1 + x)(1 + x^2)^5 \]
This is done by factoring out \( (1 + x) \) from the first four terms.

2. **Expand Each Factor**:
We need to expand both \( (1 + x)^5 \) and \( (1 + x^2)^5 \) using the Binomial Theorem.
- For \( (1 + x)^5 \):
\[ (1 + x)^5 = \sum_{k=0}^{5} \binom{5}{k} x^k \]
- For \( (1 + x^2)^5 \):
\[ (1 + x^2)^5 = \sum_{j=0}^{5} \binom{5}{j} (x^2)^j = \sum_{j=0}^{5} \binom{5}{j} x^{2j} \]

3. **Identify the Coefficient of \( x^{10} \)**:
We want to find the coefficient of \( x^{10} \) in the product of these two expansions:
\[ (1 + x)^5 \cdot (1 + x^2)^5 \]
We need to consider combinations of terms from both expansions that will yield \( x^{10} \).

4. **Possible Combinations**:
- From \( (1 + x)^5 \), we can take \( x^k \) where \( k \) can be \( 0, 2, 4, 6, 8, 10 \).
- From \( (1 + x^2)^5 \), we can take \( x^{2j} \) where \( 2j \) must complement \( k \) such that \( k + 2j = 10 \).

The combinations are:
- \( k = 0 \) and \( 2j = 10 \) (i.e., \( j = 5 \))
- \( k = 2 \) and \( 2j = 8 \) (i.e., \( j = 4 \))
- \( k = 4 \) and \( 2j = 6 \) (i.e., \( j = 3 \))
- \( k = 6 \) and \( 2j = 4 \) (i.e., \( j = 2 \))
- \( k = 8 \) and \( 2j = 2 \) (i.e., \( j = 1 \))
- \( k = 10 \) and \( 2j = 0 \) (i.e., \( j = 0 \))

5. **Calculate Each Combination**:
- For \( k = 0, j = 5 \): Coefficient = \( \binom{5}{0} \cdot \binom{5}{5} = 1 \cdot 1 = 1 \)
- For \( k = 2, j = 4 \): Coefficient = \( \binom{5}{2} \cdot \binom{5}{4} = 10 \cdot 5 = 50 \)
- For \( k = 4, j = 3 \): Coefficient = \( \binom{5}{4} \cdot \binom{5}{3} = 5 \cdot 10 = 50 \)
- For \( k = 6, j = 2 \): Coefficient = \( \binom{5}{6} \cdot \binom{5}{2} = 0 \) (not possible)
- For \( k = 8, j = 1 \): Coefficient = \( \binom{5}{8} \cdot \binom{5}{1} = 0 \) (not possible)
- For \( k = 10, j = 0 \): Coefficient = \( \binom{5}{10} \cdot \binom{5}{0} = 0 \) (not possible)

6. **Sum the Coefficients**:
\[ a_{10} = 1 + 50 + 50 = 101 \]

### Final Answer:
Thus, the coefficient \( a_{10} \) is \( 101 \).