The coefficient of `x^(8)` in the expansion of `1+(1+x) +(1+x)^(2) +…+ (1+x)^(n) (n ge 8)` is

Options

A. 1
B. 2
C. `""^(n+1)C_(n-8)`
D. `""^(n)C_(n-8)`

Answer

To find the coefficient of \( x^8 \) in the expansion of \( 1 + (1+x) + (1+x)^2 + \ldots + (1+x)^n \) where \( n \geq 8 \), we can follow these steps:

### Step 1: Identify the series
The expression can be rewritten as a geometric series:
\[ S = 1 + (1+x) + (1+x)^2 + \ldots + (1+x)^n \]
This is a geometric series with the first term \( a = 1 \) and the common ratio \( r = (1+x) \).

### Step 2: Use the formula for the sum of a geometric series
The sum of the first \( n+1 \) terms of a geometric series can be given by:
\[ S = \frac{a(1 - r^{n+1})}{1 - r} = \frac{1 - (1+x)^{n+1}}{1 - (1+x)} = \frac{1 - (1+x)^{n+1}}{-x} \]
Thus, we have:
\[ S = \frac{1 - (1+x)^{n+1}}{-x} \]

### Step 3: Simplify the expression
Rearranging gives:
\[ S = \frac{(1+x)^{n+1} - 1}{x} \]

### Step 4: Expand \( (1+x)^{n+1} \)
Using the Binomial Theorem, we can expand \( (1+x)^{n+1} \):
\[ (1+x)^{n+1} = \sum_{k=0}^{n+1} \binom{n+1}{k} x^k \]

### Step 5: Substitute back into the sum
Substituting this expansion back into our expression for \( S \):
\[ S = \frac{\sum_{k=0}^{n+1} \binom{n+1}{k} x^k - 1}{x} \]
This simplifies to:
\[ S = \sum_{k=1}^{n+1} \binom{n+1}{k} x^{k-1} \]

### Step 6: Find the coefficient of \( x^8 \)
To find the coefficient of \( x^8 \) in \( S \), we need to find the coefficient of \( x^9 \) in \( (1+x)^{n+1} \):
\[ \text{Coefficient of } x^9 = \binom{n+1}{9} \]

### Conclusion
Thus, the coefficient of \( x^8 \) in the expansion of \( 1 + (1+x) + (1+x)^2 + \ldots + (1+x)^n \) is:
\[ \boxed{\binom{n+1}{9}} \]