MATHEMATICAL INDUCTION

👨🏻‍🏫 學術台數學

[匿名]

2008-05-18 7:52 pm

Use MI to prove

2^n >= 2n

THANKS~

回答 (1)

Adelmen

2008-05-18 8:11 pm

✔ 最佳答案

When n=1
L.H.S. = 2^1 =2 = (2x1) =R.H.S.
Assume n=k is true, i.e., 2^k >= 2k
When n=k+1
L.H.S = 2^(k+1) = 2 x 2^k >= 2 x 2k =4k = 2k + 2k >= 2k + 2 = 2(k+1) = R.H.S.

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