C2n2+C2n4+…+C2n2k+…+C2n2n的值为( )
A.2n
B.22n-1
C.2n-1
D.22n-1-1
【答案】分析:根据C2n+C2n2+C2n4+…+C2n2k+…+C2n2n =C2n1+C2n3+…+C2n2n-1=22n-1 ,及C2n=1,可得所求的式子的值.
解答:解:由于C2n+C2n2+C2n4+…+C2n2k+…+C2n2n =22n-1,C2n=1,
故C2n2+C2n4+…+C2n2k+…+C2n2n =22n-1 -1,
故选:D.
点评:本题主要考查二项式系数的性质,利用了 C2n+C2n2+C2n4+…+C2n2k+…+C2n2n =C2n1+C2n3+…+C2n2n-1=22n-1 .