Animated Solution for Mathematics - Binomial Theorem: If the $1011^{th}$ term from the end in the binomial expansion of $(\frac{4x}{5}-\frac{5}{2x}{)}^{2022}$ is 1024 times $1011^{th}$ term from the beginning, then $32|x|$ is equal to

A Journey Through $2022$ Terms

Imagine you are standing before a massive, daunting wall of numbers. You are given the binomial expansion of $(\frac{4x}{5}-\frac{5}{2x}{)}^{2022}$.

At first glance, the exponent $2022$ feels like a mountain. You might be tempted to think, "How can I possibly handle two thousand and twenty-three terms?"

But here is the secret of the JEE Advanced: the problem is not testing your ability to calculate massive numbers; it is testing your ability to see the underlying structure. The binomial theorem is not just a formula; it is a mirror. Let us walk through this together.

Counting from the End

When we talk about the $1011^{th}$ term from the end, we are essentially looking at the expansion from the other side. Instead of expanding the whole thing, we use a powerful logical bridge.

The $k^{th}$ term from the end is identical to the $(n-k+2{)}^{th}$ term from the beginning. With $n=2022$ and $k=1011$, we calculate the position from the start: $2022-1011+2=1013$.

So, the $1011^{th}$ term from the end is simply $T_{1013}$. We have successfully translated a "reverse" problem into a standard "forward" problem.

Our Mathematical Toolkit

Now, we invoke the general term formula: $T_{r+1}=\left(\genfrac{}{}{0ex}{}{n}{r}\right)a^{n-r}{b}^r$. Here, $a=\frac{4x}{5}$ and $b=-\frac{5}{2x}$.

For $T_{1011}$, we set $r=1010$. For $T_{1013}$, we set $r=1012$. Notice the "minus one" rule—the index $r$ is always one less than the term number.

We write out our terms:

Notice that because $1010$ and $1012$ are even, the negative sign inside the second bracket simply evaporates. It is like magic, but it is just algebra.

The problem states that $T_{1013}=1024\cdot T_{1011}$. When we set these two expressions equal, we see the beauty of binomial symmetry.

We have $\left(\genfrac{}{}{0ex}{}{2022}{1012}\right)$ on the left and $\left(\genfrac{}{}{0ex}{}{2022}{1010}\right)$ on the right. Because $1012+1010=2022$, the symmetry property $\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\left(\genfrac{}{}{0ex}{}{n}{n-r}\right)$ guarantees that these coefficients are exactly the same.

They cancel out, leaving us with only the variable parts. This is the moment where the complexity collapses into simplicity.

Isolating $x$

We are left with:

By dividing both sides by the smaller powers, we reduce this to:

Expanding the squares, we get $\frac{25}{4x^2}=1024\cdot \frac{16x^2}{25}$. Cross-multiplying gives us:

Taking the square root twice, we find $|x|=\frac{5}{2^4}=\frac{5}{16}$. The final step is to find $32|x|$, which is $32\cdot \frac{5}{16}=10$.

You have conquered the mountain! The final answer is 10.