n∈N,試比較2^n與(n+1)^2的大小,并證明,用數(shù)學(xué)歸納法

數(shù)學(xué)人氣：852 ℃時(shí)間：2020-06-24 03:10:04

優(yōu)質(zhì)解答

正整數(shù)永遠(yuǎn)左邊大.
n=1時(shí) 左邊大3
n=2時(shí) 左邊大2
n=3時(shí) 左邊大1
當(dāng)n>=4時(shí),左右兩邊的增量分別是
[ 2^(n+1)+2 ] - [ 2^n+2 ] = 2^n
(n+1)^2 - n^2 = 2n + 1
n=4時(shí),2^n > 2n + 1
2^n = 4,2^(n+1) = 2^n + 4
2n + 1 = 5,2(n+1) + 1 = (2n + 1) + 2
n>4時(shí),
2^(n+1) > 2^n + 4
2(n+1) + 1 = (2n + 1) + 2
所以一直有 2^n > 2n+1
所以一直有2^n+2 > n^2