The coefficient of `x^(p)" in"(1+x)^(p)+(1+x)^(p+1)+…+(1+x)^(n), p lt n`, is

Options

A. `""^(n+1)C_(p+1)`
B. `""^(n-1)C_(p-1)`
C. `""^(n)C_(p)`
D. `""^(n)C_(p+1)`

Answer

To find the coefficient of \( x^p \) in the expression \( (1+x)^p + (1+x)^{p+1} + \ldots + (1+x)^n \), we can follow these steps:

### Step-by-step Solution:

1. **Understand the Binomial Expansion**:
The coefficient of \( x^p \) in \( (1+x)^k \) is given by \( \binom{k}{p} \). This means that for each term in our expression, we can find the coefficient of \( x^p \) using this formula.

2. **Identify the Range of k**:
The expression \( (1+x)^p + (1+x)^{p+1} + \ldots + (1+x)^n \) consists of terms where \( k \) takes values from \( p \) to \( n \).

3. **Sum the Coefficients**:
We need to sum the coefficients of \( x^p \) from \( (1+x)^p \) to \( (1+x)^n \):
\[ \text{Coefficient of } x^p = \sum_{k=p}^{n} \binom{k}{p} \]

4. **Use the Hockey Stick Identity**:
The Hockey Stick Identity in combinatorics states that:
\[ \sum_{i=r}^{n} \binom{i}{r} = \binom{n+1}{r+1} \]
Applying this identity with \( r = p \) and \( n = n \), we have:
\[ \sum_{k=p}^{n} \binom{k}{p} = \binom{n+1}{p+1} \]

5. **Final Result**:
Therefore, the coefficient of \( x^p \) in the given expression is:
\[ \binom{n+1}{p+1} \]

### Summary:
The coefficient of \( x^p \) in the expression \( (1+x)^p + (1+x)^{p+1} + \ldots + (1+x)^n \) is \( \binom{n+1}{p+1} \).