问题 选择题

C2n2+C2n4+…+C2n2k+…+C2n2n的值为(  )

A.2n

B.22n-1

C.2n-1

D.22n-1-1

答案

由于C2n0+C2n2+C2n4+…+C2n2k+…+C2n2n =22n-1,C2n0=1,

故C2n2+C2n4+…+C2n2k+…+C2n2n =22n-1 -1,

故选:D.

判断题
问答题 简答题