The coefficient of `t^50` in `(1+t)^41 (1-t+t^2)^40` is equal to

Answer

To find the coefficient of \( t^{50} \) in the expression \( (1+t)^{41} (1-t+t^2)^{40} \), we can break down the problem step by step.

### Step 1: Expand the expression
We start with the expression:
\[ (1+t)^{41} (1-t+t^2)^{40} \]

### Step 2: Analyze \( (1-t+t^2)^{40} \)
To analyze \( (1-t+t^2)^{40} \), we can use the binomial theorem. The term \( (1-t+t^2)^{40} \) can be expanded as:
\[ \sum_{k=0}^{40} \binom{40}{k} (1)^{40-k} (-t+t^2)^k \]
This simplifies to:
\[ \sum_{k=0}^{40} \binom{40}{k} (-1)^k t^k (1+t)^{2k} \]

### Step 3: Combine the expansions
Now we combine this with \( (1+t)^{41} \):
\[ (1+t)^{41} \sum_{k=0}^{40} \binom{40}{k} (-1)^k t^k (1+t)^{2k} \]
This can be rewritten as:
\[ \sum_{k=0}^{40} \binom{40}{k} (-1)^k t^k (1+t)^{41 + 2k} \]

### Step 4: Find the coefficient of \( t^{50} \)
We need to find the coefficient of \( t^{50} \) in the above expression. The term \( (1+t)^{41 + 2k} \) can be expanded using the binomial theorem:
\[ (1+t)^{41 + 2k} = \sum_{m=0}^{41 + 2k} \binom{41 + 2k}{m} t^m \]
Thus, we need to find the values of \( k \) and \( m \) such that:
\[ k + m = 50 \]
This implies:
\[ m = 50 - k \]
Now we need \( 50 - k \leq 41 + 2k \), which simplifies to:
\[ 50 \leq 41 + 3k \implies 9 \leq 3k \implies k \geq 3 \]
Also, since \( k \) can only go up to 40, we have:
\[ 3 \leq k \leq 40 \]

### Step 5: Check for multiples of 3
From the expansion of \( (1-t+t^2)^{40} \), we notice that the powers of \( t \) that we can obtain from \( (1-t+t^2)^{40} \) will be of the form \( 3n \) for \( n = 0, 1, 2, \ldots, 40 \). Therefore, the possible powers of \( t \) are multiples of 3.

Since \( 50 \) is not a multiple of 3, we conclude that there is no term \( t^{50} \) in the expansion.

### Final Answer
Thus, the coefficient of \( t^{50} \) in the expression \( (1+t)^{41} (1-t+t^2)^{40} \) is:
\[ \boxed{0} \]