分类：教育/科学-理工学科-数学
问题：一道数学题C(0,100)
(0,100)-C(2,100)+C(4,100)-...-C(98,100)+C(100,100)等于( )A.-2^50 B.0 C.1 D.2^50最好有过程 谢谢
回答：①(1+i)^100
=[C(0,100)-C(2,100)+C(4,100)-...-C(98,100)+C(100,100)]
+i[C(1,100)-C(3,100)+C(5,100)-...+C(97,100)-C(99,100)]；

②1+i=(√2)[cos(π/4)+isin(π/4)]，
(1+i)^100
=[(√2)^100][cos(100π/4)+isin(100π/4)]
=-2^50.

所以C(0,100)-C(2,100)+C(4,100)-...-C(98,100)+C(100,100)=-2^50.

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额外副产品
C(1,100)-C(3,100)+C(5,100)-...+C(97,100)-C(99,100)=0