N分之一加N分之二......N分之N减一

N分之一加N分之二......N分之N减一,第1张

1/n+2/n+3/n++(n-1)/n
=[1+2+3++(n-1)]/n
=(n-1)n/2/n
=(n-1)n/2n
=(n-1)/2
注 1+2+3++n=n(n+1)/2

2^(2n)-2^n=240
(2^n)^2-2^n=240
令2^n=x(x>0) (因为是指数函数,所以大于零)
所以x^2-x-240=0
(x-16)(x+15)=0
x=16 或者x=-15(舍)
所以2^n=16
n=4

1/(n+1)+1/(n+2) = 1/(2n)
2n[1/(n+1)+1/(n+2)] = 2n[1/(2n)]
2n/(n+1) + 2n/(n+2) = 1
(n+1)(n+2)[2n/(n+1) + 2n/(n+2)] = 1(n+1)(n+2)
2n(n+2) + 2n(n+1) = (n+1)(n+2)
2n^2 + 4n + 2n^2 + 2n= n^2 + 2n + n + 2
4n^2 + 6n = n^2 + 3n + 2
3n^2 + 3n - 2 = 0
n = [-b +/- (b^2 - 4ac)^05] / 2b
n = 022871, n = -07287
所以n得最小值是 -07287

考虑
|(n^2-2)/(n^2+n+1) - 1|
=| (n^2-2-n^2-n-1)/(n^2+n+1) |
=| (-n-3)/(n^2+n+1) |
=(n+3)/(n^2+n+1)
<(n+3)/n^2 (因为n^2+n+1>n^2)
限制n>3
<2n/n^2
=2/n
对任意ε>0,取N=max{2/ε,3}>0,
当n>N,就有|(n^2-2)/(n^2+n+1) - 1|<ε
根据定义,
lim (n^2-2)/(n^2+n+1) = 1
有不懂欢迎追问

“2n/2的n次方”是2n/(2^n)
题意求2n/(2^n)的最值
n=0时,2n/(2^n)=0
n=1时,2n/(2^n)=1
n=2时,2n/(2^n)=1
n=3时,2n/(2^n)=6/8=3/4
n=4时,2n/(2^n)=8/16=1/2
n=5时,2n/(2^n)=10/32=5/16

n=k时,2n/(2^n)=2k/(2^k)
以后n=k+1时,2(k+1)/[2^(k+1)]
观察上面数列知,n小于3时,数列没有规律变化。当n=3开始,数列出现递减规律,用数学归纳法,现在比较n=k时,和n=k+1时,2k/(2^k)和2(k+1)/[2^(k+1)]的大小
因为2k/(2^k)-2(k+1)/[2^(k+1)]=2k/(2^k)-(2k+2)/(22k)=(4k-2k-2)/(22k)=(2k-2)/(22k),因为k大于2,所以)(2k-2)/(22k)大于0
故2k/(2^k)大于2(k+1)/[2^(k+1)]
也即n=k+1时的项小于前面n=k的项,直到n趋近无穷大时,2n/(2^n)趋近0
总和上面结果知n大于0时,2n/(2^n)的最大值1
同理可得当nx小于0时,2n/(2^n)有最小值1


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