Coefficient of `x^(50)` in the expansion of `(1 + x)^(41) (1 - x + x^(2))^(40)` is

Options

A. 0
B. 1
C. `""^(40)C_(19)`
D. `""^(40)C_(29)`

Answer

To find the coefficient of \( x^{50} \) in the expansion of \( (1 + x)^{41} (1 - x + x^2)^{40} \), we will break down the problem into manageable steps.

### Step 1: Expand \( (1 + x)^{41} \)

The binomial expansion of \( (1 + x)^{41} \) can be expressed as:
\[ \sum_{r=0}^{41} \binom{41}{r} x^r \]
where \( \binom{41}{r} \) is the binomial coefficient representing the number of ways to choose \( r \) items from 41.

### Step 2: Expand \( (1 - x + x^2)^{40} \)

Next, we need to expand \( (1 - x + x^2)^{40} \). This can be treated as a trinomial expansion. The general term in the expansion of \( (1 - x + x^2)^{40} \) can be expressed using the multinomial theorem:
\[ \frac{40!}{a! b! c!} (1)^a (-x)^b (x^2)^c \]
where \( a + b + c = 40 \). The term will be:
\[ \frac{40!}{a! b! c!} (-1)^b x^{b + 2c} \]
Here, \( b + 2c \) represents the power of \( x \).

### Step 3: Find combinations for \( x^{50} \)

We need to find combinations of \( r \) from \( (1 + x)^{41} \) and \( b + 2c \) from \( (1 - x + x^2)^{40} \) such that:
\[ r + (b + 2c) = 50 \]
This can be rearranged to:
\[ b + 2c = 50 - r \]

### Step 4: Determine valid values of \( r \)

Since \( r \) can take values from 0 to 41, we need \( 50 - r \) to be non-negative. Therefore, \( r \) must be at most 50, but since \( r \) can only go up to 41, we have:
\[ 0 \leq r \leq 41 \]

### Step 5: Analyze \( b + 2c = 50 - r \)

Now, for each valid \( r \), we need to find non-negative integers \( b \) and \( c \) such that:
\[ b + 2c = 50 - r \]
We can express \( b \) in terms of \( c \):
\[ b = 50 - r - 2c \]
For \( b \) to be non-negative:
\[ 50 - r - 2c \geq 0 \implies c \leq \frac{50 - r}{2} \]

### Step 6: Calculate the coefficient

The coefficient of \( x^{50} \) in the expansion can now be calculated by summing over all valid \( r \):
\[ \text{Coefficient} = \sum_{r=0}^{41} \binom{41}{r} \sum_{c=0}^{\lfloor (50 - r)/2 \rfloor} \frac{40!}{(40 - (50 - r - 2c))! (50 - r - 2c)! c!} (-1)^{50 - r - 2c} \]

### Conclusion

This expression gives us the coefficient of \( x^{50} \) in the expansion of \( (1 + x)^{41} (1 - x + x^2)^{40} \).