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 $8$.
