The coefficient of `x^(n)` in the polynomial `(x+""^(2n+1)C_(0))(X+""^(2n+1)C_(1)) (x+""^(2n+1)C_(2))……(X+""^(2n+1)C_(n))` is

Options

A. (a)`n.2^(n-1)`
B. (b)`2^(2n)`
C. (c)`n.2^(n+1)`
D. (d)`(n+1)2^(n)`

Answer

To find the coefficient of \( x^n \) in the polynomial

\[ (x + \binom{2n+1}{0})(x + \binom{2n+1}{1})(x + \binom{2n+1}{2}) \ldots (x + \binom{2n+1}{n}), \]

we can follow these steps:

### Step 1: Understand the Polynomial Structure
The polynomial consists of \( n+1 \) factors, each of the form \( (x + \binom{2n+1}{k}) \) for \( k = 0, 1, 2, \ldots, n \).

### Step 2: Expand the Polynomial
To find the coefficient of \( x^n \), we need to consider how we can select \( x \) from \( n \) of the factors and the constant term from one factor. This means we need to choose \( n \) factors to contribute \( x \) and one factor to contribute the constant term.

### Step 3: Identify the Constant Term
The constant term from the \( k \)-th factor is \( \binom{2n+1}{k} \). Therefore, if we choose \( x \) from \( n \) factors and the constant from the \( (n+1) \)-th factor, the contribution to the coefficient of \( x^n \) will be:

\[ \sum_{k=0}^{n} \binom{2n+1}{k} \text{ (where we choose constant from the \( k \)-th factor)}. \]

### Step 4: Use the Binomial Theorem
Using the Binomial Theorem, we know that:

\[ \sum_{k=0}^{m} \binom{m}{k} = 2^m. \]

In our case, \( m = 2n + 1 \). Thus, we can write:

\[ \sum_{k=0}^{2n+1} \binom{2n+1}{k} = 2^{2n+1}. \]

### Step 5: Calculate the Coefficient of \( x^n \)
Since we are only interested in the first \( n+1 \) terms (from \( k=0 \) to \( k=n \)), we can use the symmetry of binomial coefficients:

\[ \sum_{k=0}^{n} \binom{2n+1}{k} = \sum_{k=n+1}^{2n+1} \binom{2n+1}{k} = 2^{2n+1}/2 = 2^{2n}. \]

### Step 6: Final Coefficient Calculation
Thus, the coefficient of \( x^n \) in the polynomial is:

\[ \text{Coefficient of } x^n = 2^{2n}. \]

### Conclusion
Therefore, the coefficient of \( x^n \) in the given polynomial is \( 2^{2n} \).

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