The coefficient of `x^(n)` in `(1+x)^(101) (1-x+x^(2))^(100)` is

Options

A. `3r+1`
B. `3r`
C. `3r+2`
D. none of these

Explanation

To find the coefficient of \( x^n \) in the expression \( (1+x)^{101} (1-x+x^2)^{100} \), we can follow these steps:

### Step 1: Rewrite the Expression
We start with the expression:
\[ (1+x)^{101} (1-x+x^2)^{100} \]
We can rewrite \( (1-x+x^2) \) as \( (1+x)(1+x^2) - x(1+x) \). However, a simpler approach is to use the binomial theorem directly.

### Step 2: Expand \( (1-x+x^2)^{100} \)
Using the binomial theorem, we can expand \( (1-x+x^2)^{100} \). This can be done by recognizing that:
\[ 1 - x + x^2 = (1+x)(1+x) - x(1+x) \]
However, we will directly apply the binomial theorem here.

### Step 3: Apply the Binomial Theorem
Using the binomial theorem, we can expand \( (1-x+x^2)^{100} \):
\[ (1-x+x^2)^{100} = \sum_{k=0}^{100} \binom{100}{k} (1)^{100-k} (-x+x^2)^k \]
This further expands to:
\[ = \sum_{k=0}^{100} \binom{100}{k} (-1)^k (x^k + \text{terms with } x^2) \]

### Step 4: Find Coefficients
Next, we need to find the coefficient of \( x^n \) in \( (1+x)^{101} \) and combine it with the coefficients from the expansion of \( (1-x+x^2)^{100} \).

1. The expansion of \( (1+x)^{101} \) gives:
\[ \sum_{m=0}^{101} \binom{101}{m} x^m \]
The coefficient of \( x^m \) is \( \binom{101}{m} \).

2. Now we need to find the coefficient of \( x^{n-m} \) in \( (1-x+x^2)^{100} \).

### Step 5: Combine Coefficients
The coefficient of \( x^n \) in the product \( (1+x)^{101} (1-x+x^2)^{100} \) is given by:
\[ \sum_{m=0}^{n} \binom{101}{m} \cdot \text{Coefficient of } x^{n-m} \text{ in } (1-x+x^2)^{100} \]

### Step 6: Use the Coefficient Formula
From the expansion of \( (1-x+x^2)^{100} \), we can find that:
- The coefficient of \( x^{n} \) is \( \binom{100}{\frac{n}{3}} \) if \( n \) is divisible by 3.
- The coefficient of \( x^{n-1} \) is \( \binom{100}{\frac{n-1}{3}} \) if \( n-1 \) is divisible by 3.

### Final Step: Combine Results
Thus, the coefficient of \( x^n \) in the original expression becomes:
\[ \binom{100}{\frac{n}{3}} + \binom{100}{\frac{n-1}{3}} \]

### Conclusion
The coefficient of \( x^n \) in \( (1+x)^{101} (1-x+x^2)^{100} \) is:
\[ \binom{100}{\frac{n}{3}} + \binom{100}{\frac{n-1}{3}} \]