Question

Find the coefficient of x^5y^8 in the expansion of (2x- y^2)^9

A 1200
B 4032
C 6543
D 1098

Answer 1

From binomial expansion we have, that:


`(2x-y^2)=\sum\limits_(k=0)^9\binom{9}{k}\cdot(2x)^(9-k)\cdot(-y^2)^k=\\\\\\= \dots+\binom{9}{4}\cdot(2x)^(9-4)\cdot(-y^2)^4+\dots=\\\\\\= \dots+\binom{9}{4}\cdot2^5x^5\cdot(-1)^4(y^2)^4+\dots= \dots+\binom{9}{4}\cdot2^5\cdot x^5y^8+\dots=\\\\\\=\dots+(9!)/(4!(9-4)!)\cdot32\cdot x^5y^8+\dots=\dots+(9\cdot8\cdot7\cdot6)/(4\cdot3\cdot2\cdot1)\cdot32\cdot x^5y^8+\dots=\\\\\\=\dots+126\cdot32\cdoy x^5y^8+\dots=\dots+\boxed{4032x^5y^8}+\dots`

Answer B.