Find the coefficient of `x^9` in the expansion of `(1+x)(1+x^2)(1+x^3)(1+x^4)......(1+x^100)`

Answer

To find the coefficient of \( x^9 \) in the expansion of

\[ (1+x)(1+x^2)(1+x^3)(1+x^4)\ldots(1+x^{100}), \]

we need to determine how many different combinations of the terms from the factors can be selected such that the exponents of \( x \) add up to 9.

### Step-by-Step Solution:

1. **Understanding the Problem**:
Each factor \( (1+x^k) \) contributes either 0 or \( k \) to the exponent of \( x \). We need to choose some of these \( k \) values such that their sum equals 9.

2. **Finding Combinations**:
We need to find all combinations of integers from 1 to 100 that sum to 9. The integers we can choose from are \( 1, 2, 3, \ldots, 100 \).

3. **Listing Combinations**:
We can list the combinations that sum to 9:
- \( 9 \)
- \( 8 + 1 \)
- \( 7 + 2 \)
- \( 6 + 3 \)
- \( 6 + 2 + 1 \)
- \( 5 + 4 \)
- \( 5 + 3 + 1 \)
- \( 5 + 2 + 2 \) (not valid since we can only pick each number once)
- \( 4 + 3 + 2 \)
- \( 4 + 2 + 2 + 1 \) (not valid since we can only pick each number once)
- \( 3 + 3 + 3 \) (not valid since we can only pick each number once)
- \( 4 + 4 + 1 \) (not valid since we can only pick each number once)
- \( 3 + 2 + 2 + 2 \) (not valid since we can only pick each number once)
- \( 2 + 2 + 2 + 2 + 1 \) (not valid since we can only pick each number once)

Valid combinations that sum to 9:
- \( 9 \)
- \( 8 + 1 \)
- \( 7 + 2 \)
- \( 6 + 3 \)
- \( 6 + 2 + 1 \)
- \( 5 + 4 \)
- \( 5 + 3 + 1 \)
- \( 4 + 3 + 2 \)

4. **Counting the Combinations**:
Now we count the valid combinations:
- \( 9 \) (1 way)
- \( 8 + 1 \) (1 way)
- \( 7 + 2 \) (1 way)
- \( 6 + 3 \) (1 way)
- \( 6 + 2 + 1 \) (1 way)
- \( 5 + 4 \) (1 way)
- \( 5 + 3 + 1 \) (1 way)
- \( 4 + 3 + 2 \) (1 way)

Total combinations = 8.

5. **Conclusion**:
The coefficient of \( x^9 \) in the expansion is therefore \( 8 \).