Key Concept: The Binomial Theorem and General Term

The Binomial Theorem provides a formula for expanding any power of a binomial $(a+b)^n$. For a positive integer $n$, the expansion is given by: $(a+b)^n = \sum_{r=0}^{n} {}^nC_r a^{n-r} b^r$ The $(r+1)$-th term in this expansion, often called the general term ($T_{r+1}$), is particularly useful for finding specific terms or coefficients without expanding the entire expression. It is given by: $T_{r+1} = {}^nC_r a^{n-r} b^r$ Here, ${}^nC_r = \frac{n!}{r!(n-r)!}$ is the binomial coefficient.

Problem Setup

We are asked to find and compare the coefficients of $x^p$ and $x^q$ in the expansion of ${\left( {1 + x} \right)^{p + q}}$. Let's identify the components of our given binomial expression with the general form $(a+b)^n$:

• $a = 1$
• $b = x$
• $n = p+q$ (This is the overall power of the binomial)

Step-by-Step Solution

1. Determine the General Term of the Expansion

Our first step is to write down the general term $T_{r+1}$ for ${\left( {1 + x} \right)^{p + q}}$. This will allow us to easily extract the coefficient of any desired power of $x$.

Using the general term formula $T_{r+1} = {}^nC_r a^{n-r} b^r$ with $n = p+q$, $a=1$, and $b=x$: $T_{r+1} = {}^{p+q}C_r (1)^{(p+q)-r} (x)^r$ Since $1$ raised to any power is $1$, the term $(1)^{(p+q)-r}$ simplifies to $1$. So, the general term becomes: $T_{r+1} = {}^{p+q}C_r x^r$ This means that the coefficient of $x^r$ in the expansion is ${}^{p+q}C_r$.

2. Find the Coefficient of $x^p$

To find the coefficient of $x^p$, we need to determine which term in the expansion contains $x^p$. From our general term $T_{r+1} = {}^{p+q}C_r x^r$, the power of $x$ is $r$. Therefore, to get $x^p$, we must set $r=p$. Substituting $r=p$ into the general term, the $(p+1)$-th term is: $T_{p+1} = {}^{p+q}C_p x^p$ The coefficient of $x^p$ is the numerical part of this term, which is ${}^{p+q}C_p$.

3. Find the Coefficient of $x^q$

Similarly, to find the coefficient of $x^q$, we need the term where the power of $x$ is $q$. Using the general term $T_{r+1} = {}^{p+q}C_r x^r$, we set $r=q$. Substituting $r=q$ into the general term, the $(q+1)$-th term is: $T_{q+1} = {}^{p+q}C_q x^q$ The coefficient of $x^q$ is the numerical part of this term, which is ${}^{p+q}C_q$.

4. Compare the Coefficients Using a Binomial Property

We now have the two coefficients we need to compare:

• Coefficient of $x^p = {}^{p+q}C_p$
• Coefficient of $x^q = {}^{p+q}C_q$

To compare these, we recall a fundamental property of binomial coefficients: Identity: ${}^nC_r = {}^nC_{n-r}$ This identity highlights the symmetry of binomial coefficients. It means that choosing $r$ items from a set of $n$ is equivalent to choosing the $n-r$ items that are left out.

Let's apply this identity to the coefficient of $x^q$. Here, $n = p+q$ (the total power of the binomial) and $r=q$: ${}^{p+q}C_q = {}^{p+q}C_{(p+q)-q}$ Now, simplify the subscript $(p+q)-q$: $(p+q)-q = p$ So, substituting this back, we get: ${}^{p+q}C_q = {}^{p+q}C_p$ This result directly shows that the coefficient of $x^q$ is equal to the coefficient of $x^p$.

Conclusion

Since ${}^{p+q}C_p = {}^{p+q}C_q$, the coefficients of $x^p$ and $x^q$ in the expansion of ${\left( {1 + x} \right)^{p + q}}$ are equal.

Key Takeaways and JEE Tips

• General Term is Your Foundation: Always start by writing the general term. It simplifies finding any specific term or coefficient.
• Identify $n$, $a$, and $b$ Correctly: In $(a+b)^n$, make sure you correctly identify $a$, $b$, and $n$ for your given problem. Here, $n=p+q$, $a=1$, $b=x$.
• Symmetry of Binomial Coefficients ($^nC_r = {}^nC_{n-r}$): This is a very common and powerful property used in Binomial Theorem problems. Recognize when to apply it. In expansions of $(1+x)^n$, the coefficient of $x^r$ is $^nC_r$, and the coefficient of $x^{n-r}$ is $^nC_{n-r}$, which are always equal. This problem is a direct application where $n=p+q$, $r=p$, and $n-r = (p+q)-p = q$.
• Distinguish $r$ from $p$ or $q$: In the general term $T_{r+1} = {}^nC_r x^r$, $r$ is a variable index. $p$ and $q$ are specific, fixed powers of $x$ that we are interested in.

The final answer is $\boxed{\text{equal}}$.