摘要:(Ⅱ)证明,
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证明:(1)
2k
=3n(n∈N);
(2)2C2n0+C2n1+2C2n2+C2n3+…+C2n2n-1+2C2n2n=3•22n-1(n∈N);
(3)2<(1+
)n<3(n∈N).
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n |
k=0 |
C | k n |
(2)2C2n0+C2n1+2C2n2+C2n3+…+C2n2n-1+2C2n2n=3•22n-1(n∈N);
(3)2<(1+
1 |
n |