Animated Solution for Mathematics - Binomial Theorem: If $n$ is a positive integer , then $(\sqrt{3}+1{)}^{2n}-(\sqrt{3}-1{)}^{2n}$ is

Imagine standing before a massive, intimidating expression: $(\sqrt{3}+1{)}^{2n}-(\sqrt{3}-1{)}^{2n}$. It looks like a mountain of powers and roots designed to induce panic.

However, the secret of the JEE Advanced is that complex-looking problems are often just elegant dances of symmetry. Let us break this down systematically.

First, we must strip away the fear. We see $\sqrt{3}$ and $1$ repeated, so let us define $x=\sqrt{3}$ and $y=1$.

Suddenly, our expression transforms into the following form:

We are no longer dealing with messy roots; we are dealing with the pure, structural beauty of the Binomial Theorem. By replacing the numbers with variables, we reveal the skeleton of the problem.

Recall the Binomial Theorem. We know that $(x+y{)}^{2n}=T_1+T_2+T_3+T_4+\dots$, where $T_k$ represents the terms of the expansion.

Now, consider $(x-y{)}^{2n}$. Because of the negative sign, the expansion becomes $T_1-T_2+T_3-T_4+\dots$.

When we subtract the second expansion from the first, the magic happens. The terms $T_1,T_3,T_5,\dots$ (the odd-positioned terms) have the same sign in both expansions and vanish into thin air:

What is left behind are the terms with opposite signs. Specifically, $T_2-(-T_2)$ becomes $2T_2$, and $T_4-(-T_4)$ becomes $2T_4$.

Our expression simplifies beautifully to:

Now, let us bring back our original values, $x=\sqrt{3}$ and $y=1$. The general term $T_{k+1}$ is given by $\left(\genfrac{}{}{0ex}{}{2n}{k}\right)x^{2n-k}{y}^k$.

Substituting our values, the second term $T_2$ is $\left(\genfrac{}{}{0ex}{}{2n}{1}\right)(\sqrt{3}{)}^{2n-1}(1{)}^1$. Notice that the power of $\sqrt{3}$ is $2n-1$, which is odd.

Because every term in our sum will have an odd power of $\sqrt{3}$, we can factor out one $\sqrt{3}$ from the entire bracket:

Look inside the bracket. Since $2n-2,2n-4,\dots$ are all even, $(\sqrt{3}{)}^{\text{even}}$ becomes an integer. Since the binomial coefficients $\left(\genfrac{}{}{0ex}{}{2n}{k}\right)$ are also integers, the entire sum inside the bracket is just one big integer, $I$.

Our expression is now $E=2\sqrt{3}\times I$. Since the product of a non-zero rational number and an irrational number is always irrational, we have conquered the mountain.