8.將f(x) = x100 + 2019除以x2 − 2x + 1可得餘式 。 

詳解 (共 1 筆)

詳解 提供者：Chen Chen

x2 − 2x + 1=(x-1)2

x100=[(x-1)+1]100=C1000(x-1)100+.....+C10099(x-1)+C100100(x-1)0

餘式為C10099(x-1)+C100100(x-1)0+2019=100(x-1)+1+2019=100x+1920