Find the coefficient of `x^r` in the expansion of `1 +(1 + x) + (1 + x)^2 +.......+ (1 + x)^n`.

Options

A. `""^(n)C_(r)`
B. `""^(n+1)C_(r )`
C. `""^(n+1)C_(r+1)`
D. none of these

Explanation

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

### Step 1: Identify the series as a geometric series
The given series can be recognized as a geometric series where the first term \( a = 1 \) and the common ratio \( r = (1 + x) \). The number of terms in the series is \( n + 1 \).

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

### Step 3: Expand \( (1 + x)^{n + 1} \) using the Binomial Theorem
According to the Binomial Theorem, we can expand \( (1 + x)^{n + 1} \) as:
\[ (1 + x)^{n + 1} = \sum_{k=0}^{n + 1} \binom{n + 1}{k} x^k \]
Thus, we have:
\[ (1 + x)^{n + 1} - 1 = \sum_{k=1}^{n + 1} \binom{n + 1}{k} x^k \]

### Step 4: Substitute back into the sum expression
Now substituting this back into our expression for \( S \):
\[ S = \frac{\sum_{k=1}^{n + 1} \binom{n + 1}{k} x^k}{x} = \sum_{k=1}^{n + 1} \binom{n + 1}{k} x^{k-1} \]

### Step 5: Change the index of summation
To find the coefficient of \( x^r \), we can change the index of summation by letting \( j = k - 1 \). Then, when \( k = 1 \), \( j = 0 \) and when \( k = n + 1 \), \( j = n \):
\[ S = \sum_{j=0}^{n} \binom{n + 1}{j + 1} x^j \]

### Step 6: Identify the coefficient of \( x^r \)
The coefficient of \( x^r \) in this series is given by \( \binom{n + 1}{r + 1} \).

### Final Answer
Thus, the coefficient of \( x^r \) in the expansion of the series \( 1 + (1 + x) + (1 + x)^2 + \ldots + (1 + x)^n \) is:
\[ \boxed{\binom{n + 1}{r + 1}} \]