Let `a_(r)` denote the coefficient of `x^(r)` in the expansion of `( 1 + x)^(P + q)`, then

Options

A. `a_(p)=a_(q)`
B. `a_(p)=-a_(q)`
C. `a_(p)a_(q)=1`
D. none of these

Answer

To solve the problem, we need to find the coefficients \( a_r \) of \( x^r \) in the expansion of \( (1 + x)^{p + q} \) and establish a relationship between \( a_p \) and \( a_q \).

### Step-by-Step Solution:

1. **Understanding the Binomial Expansion**:
The binomial theorem states that:
\[ (1 + x)^n = \sum_{r=0}^{n} \binom{n}{r} x^r \]
where \( \binom{n}{r} \) is the binomial coefficient representing the coefficient of \( x^r \).

2. **Finding the Coefficient \( a_r \)**:
In our case, we are interested in the expansion of \( (1 + x)^{p + q} \). The coefficient of \( x^r \) in this expansion is given by:
\[ a_r = \binom{p + q}{r} \]

3. **Finding \( a_p \)**:
The coefficient \( a_p \) corresponds to \( r = p \). Thus:
\[ a_p = \binom{p + q}{p} \]

4. **Finding \( a_q \)**:
Similarly, the coefficient \( a_q \) corresponds to \( r = q \). Thus:
\[ a_q = \binom{p + q}{q} \]

5. **Using the Property of Binomial Coefficients**:
We know from the property of binomial coefficients that:
\[ \binom{n}{k} = \binom{n}{n-k} \]
Applying this property here, we have:
\[ a_p = \binom{p + q}{p} = \binom{p + q}{q} = a_q \]

6. **Conclusion**:
Therefore, we conclude that:
\[ a_p = a_q \]

### Final Result:
The coefficients \( a_p \) and \( a_q \) are equal, i.e., \( a_p = a_q \).

---