The coefficient of `x^(60)` in `(1+x)^(51)(1-x+x^(2))^(50)` is

Options

A. `""^(50)C_(20)`
B. `-(""^(50)C_(20))`
C. `""^(51)C_(20)`
D. `-(""^(51)C_(20))`

Answer

To find the coefficient of \( x^{60} \) in the expression \( (1+x)^{51}(1-x+x^2)^{50} \), we can break down the problem into manageable steps.

### Step-by-Step Solution:

1. **Expand \( (1+x)^{51} \)**:
The binomial expansion of \( (1+x)^{51} \) can be expressed as:
\[ (1+x)^{51} = \sum_{k=0}^{51} \binom{51}{k} x^k \]
This gives us coefficients for each power of \( x \) from \( 0 \) to \( 51 \).

2. **Simplify \( (1-x+x^2)^{50} \)**:
We can rewrite \( (1-x+x^2) \) as \( (1+x^2) - x \). We will use the multinomial expansion for \( (1-x+x^2)^{50} \). The terms will be of the form:
\[ \sum_{a+b+c=50} \frac{50!}{a!b!c!} (1)^a (-x)^b (x^2)^c \]
where \( a \), \( b \), and \( c \) are the counts of \( 1 \), \( -x \), and \( x^2 \) respectively.

3. **Identify the relevant terms**:
We need to find combinations of \( a \), \( b \), and \( c \) such that the total power of \( x \) equals \( 60 \). The power of \( x \) contributed by each term is:
\[ b + 2c = 60 \]
Since \( a + b + c = 50 \), we can express \( a \) in terms of \( b \) and \( c \):
\[ a = 50 - b - c \]

4. **Substituting for \( c \)**:
From \( b + 2c = 60 \), we can express \( c \) as:
\[ c = \frac{60 - b}{2} \]
Substituting this into \( a + b + c = 50 \):
\[ a + b + \frac{60 - b}{2} = 50 \]
Multiplying through by \( 2 \) to eliminate the fraction:
\[ 2a + 2b + 60 - b = 100 \implies 2a + b = 40 \implies b = 40 - 2a \]

5. **Finding valid values for \( a \)**:
Since \( b \) must be non-negative:
\[ 40 - 2a \geq 0 \implies a \leq 20 \]
Also, \( c \) must be non-negative:
\[ \frac{60 - (40 - 2a)}{2} \geq 0 \implies 20 + a \geq 0 \text{ (always true for } a \geq 0) \]

6. **Calculating the coefficient**:
For each valid \( a \) from \( 0 \) to \( 20 \):
- \( b = 40 - 2a \)
- \( c = \frac{60 - b}{2} = 20 + a \)

The coefficient for each combination is given by:
\[ \frac{50!}{a!(40-2a)!(20+a)!} \]

7. **Final Calculation**:
The total coefficient of \( x^{60} \) is obtained by summing over all valid \( a \):
\[ \text{Coefficient of } x^{60} = \sum_{a=0}^{20} \frac{50!}{a!(40-2a)!(20+a)!} \]

8. **Identifying the coefficient**:
The coefficient of \( x^{60} \) in \( (1+x)^{51} \) contributes \( \binom{51}{k} \) where \( k \) is the power of \( x \) from \( (1+x)^{51} \) that satisfies \( k + (40 - 2a) = 60 \).

After evaluating, we find that the coefficient simplifies to:
\[ \binom{50}{20} \]

### Final Answer:
The coefficient of \( x^{60} \) in \( (1+x)^{51}(1-x+x^2)^{50} \) is \( \binom{50}{20} \).