Animated Solution for Mathematics - Binomial Theorem: The sum of all those terms which are rational numbers in the expansion of $(2^{1/3}+3^{1/4}{)}^{12}$ is:

Imagine you are standing before a massive algebraic structure: $(2^{1/3}+3^{1/4}{)}^{12}$. It looks intimidating, doesn't it? A sum of two irrational roots raised to the twelfth power.

If you were to expand this using the Binomial Theorem, you would generate thirteen distinct terms. Most of these terms will be messy, filled with roots and radicals. But hidden within this chaos are 'rational islands'—terms that simplify into clean, beautiful integers.

Our Map

To navigate this expansion, we rely on the Binomial Theorem's most powerful tool: the general term formula. For any expansion $(a+b{)}^n$, the $(r+1)$-th term is given by:

In our specific case, $n=12$, $a=2^{1/3}$, and $b=3^{1/4}$. Substituting these values, we get:

Using the law of exponents, $(x^m{)}^n=x^{mn}$, we can refine this into a much more manageable form:

The Sieve

Now, look closely at that expression. For $T_{r+1}$ to be a rational number, the exponents of our prime bases, $2$ and $3$, must be integers. If the exponent is a fraction, we are stuck with a root, and the term remains irrational.

This gives us two strict conditions that $r$ must satisfy:

1. $\frac{12-r}{3}$ must be an integer. 2. $\frac{r}{4}$ must be an integer.

Since $r$ is an index in a binomial expansion of power $12$, we know that $0\le r\le 12$.

For $\frac{12-r}{3}$ to be an integer, $12-r$ must be a multiple of $3$. Since $12$ is already a multiple of $3$, $r$ itself must be a multiple of $3$. Within our range, the possible values for $r$ are $\{0,3,6,9,12\}$.

Next, for $\frac{r}{4}$ to be an integer, $r$ must be a multiple of $4$. Within the same range, the possible values for $r$ are $\{0,4,8,12\}$.

Finding the Gems

This is the moment of truth. We need values of $r$ that satisfy both conditions simultaneously.

By looking at our two sets—$\{0,3,6,9,12\}$ and $\{0,4,8,12\}$—we see that the intersection is simply $r=0$ and $r=12$. There are only two rational terms in this entire expansion of thirteen terms!

Now, we simply calculate these two terms. For $r=0$:

And for $r=12$:

Adding these together, we find the sum of all rational terms is $16+27=43$.

Look at what we have achieved. We took a daunting expression and, by applying the logic of prime exponents and divisibility, stripped away the irrational noise to find the rational core.

This is the beauty of JEE mathematics—it is not about brute-force calculation, but about finding the elegant path through the forest. Keep this clarity of thought, and no problem will ever be too complex for you.