Animated Solution for Mathematics - Binomial Theorem: Coefficient of $t^{24}$ in $(1+t^2{)}^{12}(1+t^{12})(1+t^{24})$ is

Welcome, future engineer! Today, we are going to tackle a problem that might look intimidating at first glance, but beneath its complex exterior lies a beautiful, elegant structure.

We are tasked with finding the coefficient of $t^{24}$ in the expansion of:

Simplifying the Landscape

When you see a product of multiple brackets, your first instinct might be to expand everything. Resist that urge! In JEE Advanced, time is your most precious resource.

Instead, let us look for a way to simplify the expression. We have three brackets: $(1+t^2{)}^{12}$, $(1+t^{12})$, and $(1+t^{24})$.

Let us multiply the last two brackets first:

Rearranging the terms, we get $1+t^{12}+t^{24}+t^{36}$. Now, our expression looks much cleaner:

The Power of the General Term

Now, we distribute $(1+t^2{)}^{12}$ across these four terms. We get four parts:

The last term, $(1+t^2{)}^{12}\cdot t^{36}$, is irrelevant because its lowest power is $t^{36}$, which is already greater than $t^{24}$. We can safely discard it.

To handle the remaining three parts, we use the Binomial Theorem. The general term of $(1+t^2{)}^{12}$ is:

This formula is our searchlight; it tells us exactly what power of $t$ we can extract from the binomial expansion.

The Hunt for $t^{24}$

Now, let us find the contribution of each part to the coefficient of $t^{24}$:

1. For the first part, $(1+t^2{)}^{12}\cdot 1$, we need $t^{24}$ from the binomial part. Setting $2r=24$, we get $r=12$. The coefficient is $\left(\genfrac{}{}{0ex}{}{12}{12}\right)=1$.

2. For the second part, $(1+t^2{)}^{12}\cdot t^{12}$, we already have $t^{12}$ outside. To reach $t^{24}$, we need $t^{12}$ from the binomial part. Setting $2r=12$, we get $r=6$. The coefficient is $\left(\genfrac{}{}{0ex}{}{12}{6}\right)$.

3. For the third part, $(1+t^2{)}^{12}\cdot t^{24}$, we already have $t^{24}$ outside. We need $t^0$ from the binomial part. Setting $2r=0$, we get $r=0$. The coefficient is $\left(\genfrac{}{}{0ex}{}{12}{0}\right)=1$.

The Final Summation

Adding these contributions together, the total coefficient of $t^{24}$ is:

This matches our fourth option perfectly. By breaking the problem down into smaller, logical steps, we turned a daunting expression into a simple, satisfying result.