In the expansion of ` (1 + x) (1 + x+ x^(2)) …(1 + x + x^(2) +… +x^(2n))` , the sum of the coefficients is

Options

A. 1
B. 2n!
C. 2n!+1
D. `(2n +1)!`

Answer

To find the sum of the coefficients in the expansion of \( (1 + x)(1 + x + x^2) \cdots (1 + x + x^2 + \ldots + x^{2n}) \), we can follow these steps:

### Step 1: Understand the Expression
The given expression is a product of several factors, where each factor is of the form \( (1 + x + x^2 + \ldots + x^k) \) for \( k = 0, 1, 2, \ldots, 2n \).

### Step 2: Rewrite Each Factor
Each factor can be rewritten using the formula for the sum of a geometric series:
\[ 1 + x + x^2 + \ldots + x^k = \frac{1 - x^{k+1}}{1 - x} \]
Thus, we can express the entire product as:
\[ (1 + x)(1 + x + x^2) \cdots (1 + x + x^2 + \ldots + x^{2n}) = \frac{(1 - x^2)(1 - x^3)(1 - x^4) \cdots (1 - x^{2n+1})}{(1 - x)^{2n + 1}} \]

### Step 3: Find the Sum of the Coefficients
To find the sum of the coefficients of the polynomial, we substitute \( x = 1 \) into the expression. This gives us:
\[ \text{Sum of coefficients} = P(1) = \frac{(1 - 1^2)(1 - 1^3)(1 - 1^4) \cdots (1 - 1^{2n+1})}{(1 - 1)^{2n + 1}} \]
However, since \( (1 - 1)^{2n + 1} = 0 \), we need to evaluate the product in a different way.

### Step 4: Count the Number of Terms
Instead of calculating directly, we can count the number of terms in the expansion. The number of terms in the product \( (1 + x)(1 + x + x^2) \cdots (1 + x + x^2 + \ldots + x^{2n}) \) is equal to the number of ways to choose an exponent from each factor.

The first factor contributes \( 2 \) choices (0 or 1), the second contributes \( 3 \) choices (0, 1, or 2), and so on, up to \( (2n + 1) \) choices from the last factor. Thus, the total number of terms is:
\[ 2 \times 3 \times 4 \times \ldots \times (2n + 1) = (2n + 1)! \]

### Step 5: Conclusion
Therefore, the sum of the coefficients in the expansion is:
\[ \text{Sum of coefficients} = (2n + 1)! \]

### Final Answer
The sum of the coefficients in the expansion of \( (1 + x)(1 + x + x^2) \cdots (1 + x + x^2 + \ldots + x^{2n}) \) is \( (2n + 1)! \).