Find the coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3)...(1+x^{100})$ The coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3)...(1+x^{100})$.

I tried the following concept, how to sum 9 using 1-9 only without repetition.

1:9, 2:1+8,3:7+2, 4: 6+3, 5: 5+4,

6:1+2+6 ,7:1+3+5, 8:2+3+4

The answer is 8.

How will it be solved using Binomial theorem.
Yes we can count directly the cases that is


$x^9$
$x\cdot x^8$
$x\cdot x^2\cdot x^6$
$x\cdot x^3\cdot x^5$
$x^2\cdot x^7$
$x^2\cdot x^3\cdot x^4$
$x^3\cdot x^6$
$x^4\cdot x^5$


to obtain $8$.
Even if you use binomial series, you must consider the cases like you did. For example:
$$[x^9](1+x)(1+x^2)(1+x^3)...(1+x^{100})=\\
[x^9]\sum_{a_1=0}^1 {1\choose a_1} x^{a_1}\cdot \sum_{a_2=0}^1 {1\choose a_2} x^{2a_2}\cdot \sum_{a_3=0}^1 {1\choose a_3} x^{3a_3}\cdots \sum_{a_{100}=0}^1 {1\choose a_{100}} x^{100a_{100}};\\
a_1+2a_2+3a_3+\cdots +100a_{100}=9; a_i \in \{0,1\}.$$
The cases are:
$$a_9=1;\\
(a_1,a_8)=(1,1);\\
(a_2,a_7)=(1,1);\\
(a_3,a_6)=(1,1);\\
(a_4,a_5)=(1,1);\\
(a_1,a_2,a_6)=(1,1,1);\\
(a_1,a_3,a_5)=(1,1,1);\\
(a_2,a_3,a_4)=(1,1,1),$$
where other coefficients are zero.