Animated Solution for Mathematics - Binomial Theorem: If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}{)}^n$ is $\sqrt{6}:1$, then the third term from the beginning is:

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of radicals and indices. We are dealing with the expansion of $(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}{)}^n$.

Many students see these roots and immediately panic, trying to calculate values that aren't meant to be calculated. But in the world of JEE Advanced, we don't calculate; we manipulate. We seek the underlying structure.

The Art of Substitution

Let us begin by cleaning our workspace. The expression $(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}{)}^n$ is cluttered. Let us define $x=2^{1/4}$ and $a=3^{-1/4}$.

Suddenly, the expression becomes $(x+a{)}^n$. This is the power of abstraction. By hiding the complexity behind variables, we can focus on the logic of the binomial expansion.

The general term formula is our north star: $T_{r+1}=\left(\genfrac{}{}{0ex}{}{n}{r}\right)x^{n-r}{a}^r$. For the $5^{th}$ term from the beginning, we set $r=4$, giving us:

The Symmetry Trick

Now, here is where the magic happens. The question asks for the $5^{th}$ term from the end. Do not waste time counting backwards from $n$.

Instead, invoke the symmetry of the binomial theorem. The $k^{th}$ term from the end of $(x+a{)}^n$ is identical to the $k^{th}$ term from the beginning of $(a+x{)}^n$.

By simply swapping the positions of $x$ and $a$, we get:

This is a beautiful, elegant shortcut that saves you precious minutes in the exam hall.

The Ratio and the Cancellation

We are given the ratio $\frac{T_5}{T_5^{\prime }}=\sqrt{6}$. Let us write this out:

Look at that! The binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{4}\right)$ cancels out completely. It vanishes, leaving us with a pure algebraic relationship:

Solving for the Unknown

Now, we must evaluate the base $\frac{x}{a}$. Substituting our original values back in:

Substituting this back into our ratio equation:

Equating the exponents, we get $\frac{n-8}{4}=\frac{1}{2}$, which simplifies to $n-8=2$, yielding $n=10$. We have cracked the code.

The Victory Lap

Finally, we calculate the $3^{rd}$ term. With $n=10$ and $r=2$, we use $T_3=\left(\genfrac{}{}{0ex}{}{10}{2}\right)x^{10-2}{a}^2$.

Multiplying these together, we get $45\cdot 4\cdot \frac{1}{\sqrt{3}}=\frac{180}{\sqrt{3}}$. Rationalizing the denominator, we arrive at $60\sqrt{3}$.

Take a moment to appreciate this. We started with a terrifying expression involving fourth roots, and through the power of symmetry and substitution, we reduced it to simple arithmetic. This is the essence of JEE Advanced mathematics—not brute force, but elegant, structured thinking.