Animated Solution for Mathematics - Binomial Theorem: If the coefficients of $x^7$ in $(a{x}^2+\frac{1}{2bx}{)}^{11}$ and $x^{-7}$ in $(ax-\frac{1}{3b{x}^2}{)}^{11}$ are equal, then

Welcome, aspiring engineer! Today, we are going to dive deep into the world of binomial expansions. This problem is not just about finding a coefficient; it is about mastering the art of algebraic manipulation and recognizing the hidden symmetry in binomial coefficients.

Imagine you are standing before a massive, complex expression like $(a{x}^2+\frac{1}{2bx}{)}^{11}$. It looks intimidating, doesn't it? But remember, every complex problem is just a collection of simple, elegant steps waiting to be uncovered.

Our first goal is to find the coefficient of $x^7$. We use the general term formula:

By isolating the variable $x$, we find the exponent $22-3r$. Setting this to $7$, we find $r=5$. This is the key!

Now, we move to the second expression $(ax-\frac{1}{3b{x}^2}{)}^{11}$. We repeat the process, finding $r=6$ for $x^{-7}$. The magic happens when we equate the two coefficients.

Using the property $\left(\genfrac{}{}{0ex}{}{11}{5}\right)=\left(\genfrac{}{}{0ex}{}{11}{6}\right)$, we see the combinations vanish, leaving us with a simple algebraic equation. This is the beauty of mathematics—the most complex expressions often simplify into something elegant and manageable.

Let us break down the first expression: $(a{x}^2+\frac{1}{2bx}{)}^{11}$. We are hunting for the coefficient of $x^7$.

Using the general term formula $T_{r+1}=\left(\genfrac{}{}{0ex}{}{11}{r}\right)(a{x}^2{)}^{11-r}(\frac{1}{2bx}{)}^r$, we separate the constants from the variables. The variable part becomes:

We set $22-3r=7$, which gives $3r=15$, so $r=5$. Substituting $r=5$ back into the constant part, we get:

This is our first coefficient. It feels like we have done a lot of work, but we are only halfway there!

Now, we turn to the second expression: $(ax-\frac{1}{3b{x}^2}{)}^{11}$. We need the coefficient of $x^{-7}$.

The general term is $T_{r+1}=\left(\genfrac{}{}{0ex}{}{11}{r}\right)(ax{)}^{11-r}(-\frac{1}{3b{x}^2}{)}^r$. Again, isolating the variable $x$, we get:

We set $11-3r=-7$, which gives $3r=18$, so $r=6$. Substituting $r=6$ into the constant part, we get:

Now, the problem states that $C_1=C_2$. We equate them:

Here is where the elegance of the binomial theorem shines. Because $\left(\genfrac{}{}{0ex}{}{11}{5}\right)=\left(\genfrac{}{}{0ex}{}{11}{6}\right)$, these terms cancel out completely! We are left with:

Cross-multiplying gives $729a^6{b}^6=32a^5{b}^5$. Dividing both sides by $a^5{b}^5$, we arrive at the final, beautiful result:

You have successfully navigated the complexity and found the truth hidden within the algebra. Keep this mindset, and no JEE problem will ever be too daunting!