The coefficient of `x^(301` ub the expansion of `(1+x)^(500)+x(1+x)^(499)+x^(2)(1+x)^(498)+…+x^(500)` is

Options

A. `"^(501)C_(301)`
B. `"^(500)C_(301)`
C. `"^(501)C_(300)`
D. none of these

Explanation

To find the coefficient of \( x^{301} \) in the expansion of

\[ (1+x)^{500} + x(1+x)^{499} + x^2(1+x)^{498} + \ldots + x^{500}, \]

we can simplify the expression step by step.

### Step 1: Rewrite the Expression
The given expression can be rewritten as:

\[ (1+x)^{500} + x(1+x)^{499} + x^2(1+x)^{498} + \ldots + x^{500} = \sum_{k=0}^{500} x^k (1+x)^{500-k}. \]

### Step 2: Factor Out Common Terms
Notice that we can factor out \( (1+x)^{500} \):

\[ (1+x)^{500} \left( 1 + \frac{x(1+x)^{499}}{(1+x)^{500}} + \frac{x^2(1+x)^{498}}{(1+x)^{500}} + \ldots + \frac{x^{500}}{(1+x)^{500}} \right). \]

This simplifies to:

\[ (1+x)^{500} \left( 1 + x \cdot \frac{1}{1+x} + x^2 \cdot \frac{1}{(1+x)^2} + \ldots + x^{500} \cdot \frac{1}{(1+x)^{500}} \right). \]

### Step 3: Recognize the Geometric Series
The series inside the parentheses is a geometric series with first term \( 1 \) and common ratio \( \frac{x}{1+x} \):

\[ \text{Sum} = \frac{1 - \left(\frac{x}{1+x}\right)^{501}}{1 - \frac{x}{1+x}} = \frac{1 - \left(\frac{x}{1+x}\right)^{501}}{\frac{1}{1+x}} = (1+x) \left( 1 - \left(\frac{x}{1+x}\right)^{501} \right). \]

### Step 4: Combine the Results
Now, we have:

\[ (1+x)^{500} \cdot (1+x) \left( 1 - \left(\frac{x}{1+x}\right)^{501} \right) = (1+x)^{501} \left( 1 - \left(\frac{x}{1+x}\right)^{501} \right). \]

### Step 5: Find the Coefficient of \( x^{301} \)
Now we need to find the coefficient of \( x^{301} \) in:

\[ (1+x)^{501} - \left(\frac{x^{501}}{(1+x)^{501}}\right). \]

The coefficient of \( x^{301} \) in \( (1+x)^{501} \) is given by \( \binom{501}{301} \).

### Step 6: Coefficient from the Second Term
The second term \( \left(\frac{x^{501}}{(1+x)^{501}}\right) \) contributes no terms to \( x^{301} \) since it starts at \( x^{501} \).

### Final Result
Thus, the coefficient of \( x^{301} \) in the entire expression is:

\[ \binom{501}{301}. \]