It is given that $(1+x)(1+{x}^{2})(1+{x}^{3})(1+{x}^{4})......(1+{x}^{100})$ ,Find the coefficient of ${x}^{9}$ in the expansion.

Question

I saw a question in my textbook but i stuck in it. The question is:

It is given that $(1+x)(1+{x}^{2})(1+{x}^{3})(1+{x}^{4})......(1+{x}^{100})$

a-)Find the coefficient of ${x}^{9}$ in the expansion.

b-)Find the coefficient of ${x}^{28}$ in the expansion.

Apart from this question, i wonder that what would happen if the question were in the form of $(1+a_0x)(1+a_1{x}^{2})(1+a_2{x}^{3})(1+a_3{x}^{4})......(1+a_{99}{x}^{100})$ where $a_1....a_{99}$ are rational numbers.

a-)Find the coefficient of ${x}^{9}$ in the expansion.

b-)Find the coefficient of ${x}^{28}$ in the expansion.

I am open to hints ,shortcuts and full solution for both of these two questions.Thanks for your helps..

Answer 1

When expanding the product, you run over all subsets $S$ of $\{1,2,\ldots,100\}$, compute the product of the second summand of the $i$th factor for all $i\in S$ and the first summand for all $i\notin S$. As the first summand is always $1$, we have $$ \prod_{i=1}^{100}(1+a_ix^i) = \sum_{S}\prod_{i\in S}a_ix^i=\sum_{S}\left(\prod_{i\in S}a_i\right)x^{\sum_{i\in S}i}.$$ Now to find the coefficient of $x^k$, we need only consider all $S$ with $\sum_{i\in S}i=k$. In the original problem statement, we are thus asked to find the number of ways to write $k$ as sum of strictly increasing positive integers ($\le 100$, to be precise - but that does not play a role for $k\le 100$).

As a warm-up, note that $$9=8+1=7+2=6+3=6+2+1=5+4=5+3+1=4+3+2 $$ gives us all(?) $8$ possible such sums for $k=9$, so the answer to the first problem should be $8$, shouldn't it? In your generalization, we would instead obtain $$ a_9+a_8a_1+a_7a_2+a_6a_3+a_6a_2a_1+a_5a_4+a_5a_3a_1+a_4a_3a_2.$$

Answer 2

Hint : The coefficient of $x^9$ is the number of ways you can decompose $9$ as a sum of distincts positive integers.

Answer 3

For the first question, notice that the coefficient of $x^9$ in $(1+x)(1+x^2)\cdots(1+x^{100})$ is the same as the coefficient of $x^9$ in $P(x)=(1+x)(1+x^2)\cdots(1+x^{9})$. Now, how can you achieve $x^9$? \begin{align*}&1\cdot1\cdot1\cdot1\cdot1\cdot1\cdot1\cdot1\cdot x^9\\ &x\cdot1\cdot1\cdot1\cdot1\cdot1\cdot1\cdot x^8\cdot1\\ &1\cdot x^2\cdot1\cdot1\cdot1\cdot1\cdot x^7\cdot1\cdot1\\ &...\end{align*} I obtained $8$. Can you generalize this idea?