Animated Solution for Mathematics - Binomial Theorem: The sum of all rational terms in the expansion of $(1+2^{1/3}+3^{1/2}{)}^6$ is equal to

Mastering Multinomial Expansions

Welcome, future engineers. Today, we are going to demystify a problem that often intimidates students at first glance: the expansion of $(1+2^{1/3}+3^{1/2}{)}^6$.

When you see a trinomial raised to a power, your instinct might be to panic. You might think, "Do I have to expand this entire thing?"

The answer is a resounding no. In the world of JEE Advanced, we don't brute-force; we use the elegance of the Multinomial Theorem to surgically extract exactly what we need.

The General Term

Imagine you are standing before this massive expansion. You don't need to see the whole structure; you only need to see the individual building blocks.

The Multinomial Theorem gives us the "DNA" of any term in this expansion. The general term $T$ is given by:

Here, $\alpha ,\beta ,$ and $\gamma$ are the exponents for $1,2^{1/3},$ and $3^{1/2}$ respectively. The golden rule of this expansion is the constraint: $\alpha +\beta +\gamma =6$.

This is our universe. Every valid term in this expansion must satisfy this equation, where $\alpha ,\beta ,\gamma$ are non-negative integers.

Now, let's talk about the "Rationality Filter." We are looking for rational terms. Look at the expression again.

We have $2^{\beta /3}$ and $3^{\gamma /2}$. For a term to be rational, the exponents of these irrational bases must be integers. This means $\beta$ must be a multiple of $3$, and $\gamma$ must be a multiple of $2$.

This is where the magic happens. We have constrained our search space significantly: - For $\beta$, the possible values are $\{0,3,6\}$. - For $\gamma$, the possible values are $\{0,2,4,6\}$.

We are no longer guessing; we are hunting. We will systematically test these combinations, always keeping the constraint $\alpha =6-\beta -\gamma \ge 0$ in our back pocket as a safety net.

Case by Case

Let's begin our search. We will organize our work by the value of $\beta$.

If $\beta =0$, our term is $2^0=1$, which is rational. Now we check $\gamma \in \{0,2,4,6\}$.

- If $\gamma =0$, then $\alpha =6$. The term is:

- If $\gamma =2$, then $\alpha =4$. The term is:

- If $\gamma =4$, then $\alpha =2$. The term is:

- If $\gamma =6$, then $\alpha =0$. The term is:

If $\beta =3$, our term involves $2^{3/3}=2^1=2$, which is rational. Now we check $\gamma \in \{0,2,4,6\}$.

- If $\gamma =0$, then $\alpha =3$. The term is:

- If $\gamma =2$, then $\alpha =1$. The term is:

- If $\gamma =4$, then $\alpha =6-3-4=-1$. This is impossible! We stop here.

If $\beta =6$, our term involves $2^{6/3}=2^2=4$, which is rational.

- If $\gamma =0$, then $\alpha =0$. The term is:

We have navigated the maze. We have identified every single rational term. Now, we simply bring them together.

The sum of all rational terms is the sum of the values we just calculated:

Adding these up with care, we arrive at $612$.

Look at what you have achieved. You didn't just solve a problem; you dismantled a complex expression by understanding its fundamental constraints.

This is the essence of JEE Advanced mathematics. It is not about memorizing formulas; it is about understanding the conditions that govern the system. Keep this systematic approach in your toolkit, and no expansion will ever intimidate you again.