The coefficients of `x^(p)` and `x^(q)(p,q in N)` in the expansion of `(1+x)^(p+q)` are

Options

A. equal
B. equal with opposite sign
C. reciprocal of each other
D. None of these

Answer

To solve the problem of finding the coefficients of \(x^p\) and \(x^q\) in the expansion of \((1+x)^{p+q}\), we will follow these steps:

### Step 1: Understand the Binomial Expansion
The binomial theorem states that:
\[ (1+x)^n = \sum_{r=0}^{n} \binom{n}{r} x^r \]
where \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\).

### Step 2: Identify the Coefficients
In our case, we need to find the coefficients of \(x^p\) and \(x^q\) in the expansion of \((1+x)^{p+q}\).

- The coefficient of \(x^p\) is given by:
\[ \text{Coefficient of } x^p = \binom{p+q}{p} = \frac{(p+q)!}{p! \cdot q!} \]

- The coefficient of \(x^q\) is given by:
\[ \text{Coefficient of } x^q = \binom{p+q}{q} = \frac{(p+q)!}{q! \cdot p!} \]

### Step 3: Compare the Coefficients
From the above calculations, we can see that:
\[ \text{Coefficient of } x^p = \frac{(p+q)!}{p! \cdot q!} \]
\[ \text{Coefficient of } x^q = \frac{(p+q)!}{q! \cdot p!} \]

### Step 4: Conclusion
Since both coefficients are equal:
\[ \text{Coefficient of } x^p = \text{Coefficient of } x^q \]
Thus, we conclude that the coefficients of \(x^p\) and \(x^q\) in the expansion of \((1+x)^{p+q}\) are equal.

### Final Answer
The coefficients of \(x^p\) and \(x^q\) in the expansion of \((1+x)^{p+q}\) are equal.

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