Question

The coefficient of x9 in the expansion of (1 + x) (1 + x2) (1 + x3) .........(1+x100) is​

Answer 1

Heya mate..

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

[x9]∏n≥1(1+xn)=[x9]∏n≥011−x2n+1

[x9]∏n≥1(1+xn)=[x9]∏n≥011−x2n+1

but that is not so useful here. However, since 1+2+3+4>91+2+3+4>9, any partition of 99 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+6,1+3+5,2+3+4

1+2+6,1+3+5,2+3+4

so, simply:

[x9]∏n≥1(1+xn)=1+4+3=8.

HOPE IT HELPS UH ☺️❤️