Find the coefficient of $x^9$ in $(1+x)(1+x^2)(1+x^3)\cdots(1+x^{100})$

Question

This question had come in jee advanced 2015. Give a hint to solve it.

Answer 1

You need to find the number of ways to write $9$ as a sum of subset of numbers $\{1,2,\cdots,100\}$. For example $1+2+6=9$. Are there any others? Note that you don't need to worry about $x^k$ for $k>9$. If you want to check your work, the correct answer should be $8$ ways.

Answer 2

You have $100$ factors, each a sum of a $1$ and a positive power of $x$. When the product is expanded as a sum, every term results from picking a $1$ from some factors and a positive power of $x$ from others. If the positive powers of $x$ that you pick in one case are $x^3$, $x^{20}$, and $x^{35}$, then the term is $x^{3+20+35} = x^{58}$.

If they are $x^2$, $x^3$, and $x^4$, then the term is $x^{2+3+4}=x^9$. ${}\qquad 2+3+4=9$.

If they are $x^1$, $x^2$, and $x^6$, then the term is $x^{1+2+6} = x^9$. ${}\qquad 1+2+6=9$.

So the question is: In how many ways can you write $9$ as a sum of distinct positive integers? That will be the coefficient of $x^9$.

Answer 3

The number of partitions into distinct parts equals the number of partitions into odd parts, hence:

$$ [x^9]\prod_{n\geq 1}(1+x^n)=[x^9]\prod_{n\geq 0}\frac{1}{1-x^{2n+1}}$$ but that is not so useful here. However, since $1+2+3+4>9$, any partition of $9$ into distinct parts has at most three parts. There are just one partition in a single part and four partitions in two parts. The partitions in three parts are: $$ 1+2+6,\quad 1+3+5,\quad 2+3+4$$ so, simply: $$[x^9]\prod_{n\geq 1}(1+x^n) = 1+4+3 = \color{red}{8}.$$