Q.If $x^p$ occurs in the expansion of $\left(x^2 + \dfrac{1}{x}\right)^{2n}$, prove that its coefficient is $\dfrac{(2n)!}{\left(\dfrac{4n - p}{3}\right)!\,\left(\dfrac{2n + p}{3}\right)!}$.

The key idea here is the Binomial Theorem Expansion. We find the general term, determine the power of $x$, and then equate it to $p$ to find the corresponding value of $r$.

1. The general term $T_{r+1}$ in the expansion of $(a+b)^N$ is given by $T_{r+1} = \binom{N}{r} a^{N-r} b^r$. For $\left(x^2 + \dfrac{1}{x}\right)^{2n}$, we have $a=x^2$, $b=x^{-1}$, and $N=2n$.

2. Simplify the powers of $x$:

3. For $x^p$ to occur, the exponent of $x$ must be $p$:

4. The coefficient of $x^p$ is $\binom{2n}{r}$. Substitute the value of $r$: $\text{Coefficient} = \binom{2n}{\frac{4n - p}{3}} = \frac{(2n)!}{\left(\frac{4n - p}{3}\right)! \left(2n - \frac{4n - p}{3}\right)!}$ …

When we expand a binomial expression like $(A+B)^N$, we are essentially looking at all possible combinations of choosing $A$ or $B$ from each of the $N$ factors. The Binomial Theorem provides a systematic way to write out these terms. Each term in the expansion has a specific coefficient and a specific power of the variables involved.

To find the coefficient of a particular power of $x$, say $x^p$, we need to:

1. Identify the general term of the expansion. This term will contain $x$ raised to some power, which depends on the term's index.
2. Simplify the $x$ part of the general term to get a single power of $x$.
3. Set this power of $x$ equal to $p$ and solve for the term's index. This tells us which term contains $x^p$.
4. Substitute this index back into the coefficient part of the general term.

Let's apply this to the given problem.

1. Write the general term of the expansion.

   The given expression is $\left(x^2 + \dfrac{1}{x}\right)^{2n}$. This is in the form $(A+B)^N$, where $A = x^2$, $B = \dfrac{1}{x}$, and $N = 2n$.

   The general term, often denoted as $T_{r+1}$, in the expansion of $(A+B)^N$ is given by:

   $T_{r+1} = \binom{N}{r} A^{N-r} B^r$

   Substituting our values:

   $T_{r+1} = \binom{2n}{r} (x^2)^{2n-r} \left(\dfrac{1}{x}\right)^r$

2. Simplify the power of $x$ in the general term.

   We need to combine all the $x$ terms. Recall that $\dfrac{1}{x} = x^{-1}$.

   $T_{r+1} = \binom{2n}{r} x^{2(2n-r)} (x^{-1})^r$

   $T_{r+1} = \binom{2n}{r} x^{4n - 2r} x^{-r}$

   Using the rule $a^m \cdot a^n = a^{m+n}$:

   $T_{r+1} = \binom{2n}{r} x^{4n - 2r - r}$

   $T_{r+1} = \binom{2n}{r} x^{4n - 3r}$

3. Equate the power of $x$ to $p$ and solve for $r$.

   We are looking for the coefficient of $x^p$. So, we set the exponent of $x$ in our general term equal to $p$:

   $4n - 3r = p$

   Now, we solve for $r$:

   $3r = 4n - p$

   $r = \dfrac{4n - p}{3}$ …