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.
C2n2+C2n4+…+C2n2k+…+C2n2n的值为( )
A.2n
B.22n-1
C.2n-1
D.22n-1-1
由于C2n0+C2n2+C2n4+…+C2n2k+…+C2n2n =22n-1,C2n0=1,
故C2n2+C2n4+…+C2n2k+…+C2n2n =22n-1 -1,
故选:D.