Animated Solution for Mathematics - Binomial Theorem: The coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3)\dots (1+x^{100})$ is \dots.

Welcome, future engineer! Take a deep breath and look at this expression:

At first glance, it looks like a monster, doesn't it? A product of one hundred binomials. If you were to try and expand this using brute force, you would be here until the next century.

But in the world of JEE Advanced, we never fight the monster head-on; we outsmart it. We look for the hidden geometry, the combinatorial soul of the problem.

The Selection Mindset

Imagine you are standing in front of 100 shelves. On the first shelf, you have a box labeled '1' and a box labeled '$x^1$'. On the second shelf, you have '1' and '$x^2$'.

On the $k$-th shelf, you have '1' and '$x^k$'. To create a term in the final expansion, you must walk down the aisle and make a choice at every single shelf: either you pick the '1', or you pick the '$x^k$'.

When you multiply your choices together, the exponents add up. If you pick $x^{k_1}$ from one shelf, $x^{k_2}$ from another, and so on, the resulting term is $x^{k_1+k_2+\dots }$.

Our goal is to find the coefficient of $x^9$. This means we are looking for the number of ways to choose a set of exponents such that $k_1+k_2+\cdots =9$. This is not algebra; this is a puzzle of partitions!

The Constraint of Distinctness

Here is the 'Aha!' moment. Notice that each shelf is unique. There is only one shelf for $x^1$, one for $x^2$, and so on.

You cannot pick $x^2$ twice because there is no second shelf with an $x^2$ option. This is the crucial constraint: the exponents you choose must be distinct positive integers.

We are not just partitioning 9; we are partitioning 9 into distinct parts. This restriction simplifies our life immensely.

The Systematic Hunt

Let us hunt for these partitions systematically. We will categorize them by the number of parts we choose.

Case 1: One Part If we choose only one exponent, it must be 9 itself. There is only one way: $\{9\}$. That is 1 way.

Case 2: Two Distinct Parts We need $a+b=9$ with $a<b$. Let us list them carefully: 1. $1+8=9$ 2. $2+7=9$ 3. $3+6=9$ 4. $4+5=9$

That gives us 4 distinct pairs. We stop here because $5+4$ is just a repeat of $4+5$.

Case 3: Three Distinct Parts We need $a+b+c=9$ with $a<b<c$. Let us be methodical:

- Start with $a=1$: We need $b+c=8$. Possible pairs $(b,c)$ where $1<b<c$ are $(2,6)$ and $(3,5)$. (Note: $(4,4)$ is invalid because parts must be distinct). - Start with $a=2$: We need $b+c=7$. The only pair where $2<b<c$ is $(3,4)$. - If we try $a=3$, we need $b+c=6$. But $b$ must be greater than 3, so $b$ would have to be at least 4, making $c$ at most 2, which violates $b<c$.

So, we have 3 valid triplets: $\{1,2,6\}$, $\{1,3,5\}$, and $\{2,3,4\}$.

Case 4: Four or More Parts Can we do four parts? Let us check the minimum possible sum of 4 distinct positive integers:

Since $10>9$, it is mathematically impossible to partition 9 into 4 or more distinct parts. The well has run dry!

The Victory Lap

Now, we simply add our successes together. We have 1 way from Case 1, 4 ways from Case 2, and 3 ways from Case 3.

The total number of ways is $1+4+3=8$.

And there you have it! The coefficient of $x^9$ is 8. You didn't need to expand a 100-term polynomial. You just needed to understand the logic of the selection. Keep this mindset—always look for the constraint, always be systematic, and the most terrifying problems will crumble before you.