The coefficient of `x^(10)` in the expansion of `1+(1+x) +…+ (1+x)^(20)` is

Options

A. `""^(19) C_(9)`
B. `""^(20) C_(10)`
C. `""^(21) C_(11)`
D. `""^(22) C_(12)`

Answer

To find the coefficient of \( x^{10} \) in the expansion of \( 1 + (1+x) + (1+x)^2 + \ldots + (1+x)^{20} \), we can follow these steps:

### Step 1: Recognize the Series
The given series can be expressed as:
\[ S = 1 + (1+x) + (1+x)^2 + \ldots + (1+x)^{20} \]
This is a geometric series where 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 \( n \) terms of a geometric series can be calculated using the formula:
\[ S_n = \frac{a(r^n - 1)}{r - 1} \]
In our case, \( n = 21 \) (since we have terms from \( 0 \) to \( 20 \)), so:
\[ S = \frac{1 \cdot ((1+x)^{21} - 1)}{(1+x) - 1} = \frac{(1+x)^{21} - 1}{x} \]

### Step 3: Find the Coefficient of \( x^{10} \)
We need to find the coefficient of \( x^{10} \) in the expression:
\[ \frac{(1+x)^{21} - 1}{x} \]
This can be rewritten as:
\[ \frac{(1+x)^{21}}{x} - \frac{1}{x} \]
To find the coefficient of \( x^{10} \), we need the coefficient of \( x^{11} \) in \( (1+x)^{21} \) because dividing by \( x \) shifts the powers down by 1.

### Step 4: Use the Binomial Theorem
The coefficient of \( x^k \) in the expansion of \( (1+x)^n \) is given by \( \binom{n}{k} \). Therefore, the coefficient of \( x^{11} \) in \( (1+x)^{21} \) is:
\[ \binom{21}{11} \]

### Step 5: Final Answer
Thus, the coefficient of \( x^{10} \) in the original series is:
\[ \binom{21}{11} \]