Animated Solution for Mathematics - Binomial Theorem: The coefficient of $x^{256}$ in the expansion of $(1-x{)}^{101}(x^2+x+1{)}^{100}$ is:

Welcome, future engineers, to another masterclass in the art of problem-solving. Today, we are staring down a beast of an expression: $(1-x{)}^{101}(x^2+x+1{)}^{100}$.

When you see powers like $101$ and $100$, your instinct might be to panic or reach for a calculator, but in the JEE Advanced arena, we don't calculate; we strategize. Let us embark on this journey to find the coefficient of $x^{256}$.

The first thing that should catch your eye is the mismatch in exponents. We have $(1-x{)}^{101}$ and $(x^2+x+1{)}^{100}$.

This is a classic setup. We want to group these terms, but we cannot do that while the powers are different. So, let us perform a simple, elegant surgery.

We rewrite $(1-x{)}^{101}$ as $(1-x)\times (1-x{)}^{100}$. Now, look at what we have:

By pulling that power of $100$ outside, we have created a beautiful, symmetric structure.

Now, focus your attention on the term inside the square brackets: $(1-x)(x^2+x+1)$. If you have been practicing your algebra, your heart should skip a beat here.

This is the classic difference of cubes identity! It collapses instantly into $(1-x^3)$.

Suddenly, our terrifying expression has transformed into:

This is the power of pattern recognition. We have taken a complex multinomial and reduced it to a simple binomial expansion.

We are not done yet. We need the coefficient of $x^{256}$. Let us distribute that $(1-x)$ across the binomial expansion.

This gives us two distinct terms: $(1-x^3{)}^{100}$ and $-x(1-x^3{)}^{100}$. We must analyze these separately.

For the first part, $(1-x^3{)}^{100}$, the general term is:

We set $3r=256$. But wait, $256$ is not divisible by $3$. This means the first part contributes absolutely nothing to the coefficient of $x^{256}$. It is a ghost term!

Now, we turn to the second part: $-x(1-x^3{)}^{100}$. We need the total power to be $x^{256}$.

Since we already have an $x$ outside, we need the binomial expansion to provide $x^{255}$. So, we set $3r=255$.

Solving this gives us $r=85$. This is a valid integer!

Now, we calculate the coefficient: we have the $-1$ from the $-x$ term, multiplied by the coefficient from the binomial expansion, which is $\left(\genfrac{}{}{0ex}{}{100}{85}\right)(-1{)}^{85}$.

Since $85$ is odd, $(-1{)}^{85}$ is $-1$. Thus, we have:

We have arrived at our answer, but wait—check the options. $\left(\genfrac{}{}{0ex}{}{100}{85}\right)$ is not there. Do not panic!

Use the symmetry property $\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\left(\genfrac{}{}{0ex}{}{n}{n-r}\right)$. Therefore:

And there it is, option (b). You have successfully navigated the trap, simplified the expression, and found the hidden symmetry. This is how you conquer JEE Advanced.