题目内容
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.
故C2n2+C2n4+…+C2n2k+…+C2n2n =22n-1 -1,
故选:D.
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