设数列{an}满足a1=2,an+1=an+(1/an),(n=1,2,3…).(1)证明:an>(2n+1)1/2(根号)对一切正整数n都成立
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/09 03:00:44
xRJA ]nE`zpH}Ш2(# @3ӇwO
ޞxԗߟٳu,joܯ0P|=])A30#|l1/JH(Fe&ŖnjXwM%!E !~@;S~tG¿=G
/0~#[|t9nMUFGWWHLFR:0T0[I$&0 ZM4EU>lHc7 4P%izM]gM̝J#
YNEe-PKr([%w-<ɯۊ hAY>bL/ /vB!!WBh>@MTfq]Q jKi#-8$2{J<1!,
设数列{an}满足a1=2,an+1=an+(1/an),(n=1,2,3…).(1)证明:an>(2n+1)1/2(根号)对一切正整数n都成立
设数列{an}满足a1=2,an+1=an+(1/an),(n=1,2,3…).(1)证明:an>(2n+1)1/2(根号)对一切正整数n都成立
设数列{an}满足a1=2,an+1=an+(1/an),(n=1,2,3…).(1)证明:an>(2n+1)1/2(根号)对一切正整数n都成立
用数学归纳法证明,
当n=1时,a1=2>√(2*1+1)=√3,成立;
当n=2时,a2=2+1/2=5/2>√(2*2+1)=√5,成立;
设n=k时,原式成立,ak>√(2k+1),(ak)²>2k+1,∵a(k+1)=ak+1/(ak),∴ak*a(k+1)=(ak)²+1,∵ak>0,且ak≠a(k+1),∴(ak)²+[a(k+1)]²>2ak*a(k+1)=2(ak)²+2,[a(k+1)]²>(ak)²+2>2(k+1)+1,则a(k+1)>√[2(k+1)+1],当n=k+1时,原式成立,故an>根号下2n+1成立.
a[n+1]^2=a[n]^2+2+1/a[n]^2,
a[n+1]^2-a[n]^2=2+1/a[n]^2,...
累加,得a[n]^2-a[1]^2=2(n-1)+1/a[n-1]^2+1/a[n-2]^2+...+1/a[1]^2
设数列an满足a1=2,a(n+1)=3an+2^(n-1),求an2,设数列an满足a1=2,a(n+1)=3an+2n,求an
设数列AN满足A1等于1,3(A1+a2+~+AN)=(n+2)an,求通向公式
设数列{an}满足a1=1, an=(4an-1 +2)/(2an-1 +7) ,则通项xn=?
关于数列、等差数列的题目设数列an满足an+1=an-2且a1=241)判断an是什么数列2)若an
数列{an}满足a1=a,an+1=1+1/an.若3/2
数列an满足a1=1,a(n+1)=an/[(2an)+1],求a2010
数列{an}满足a1=2,a(n+1)=2an+n+2,求an
数列[An]满足a1=2,a(n+1)=3an-2 求an
数列{An}满足a1=1/2,a1+a2+..+an=n方an,求an
数列an满足an+1=根号(an^2+1)+an,a1=a>0,求an通项公式
设数列an满足a1=2,a(n+1)-an=3x2的2n-1次方,求数列an的通项公式
设数列{an}满足lg(1+a1+a2+...+an)=n+1,求通项公式an
已知数列{an},{bn}满足a1=2,2an=1+2an*an+1,设{bn}=an-1求数列{1n}为等差数列急!!!
已知数列{an}满足a1=1,an+1=2an+2.(1)设bn=2^n/an,求证:数列{bn}是等差数列.(2)求数列{an}的通项公式.a(n+1)
设数列{an}满足:a1+a2/2+a3/3+a4/4……+an/n=An+B,其中A、B为常数.数列{an}是否为等差数列?
设{an}是a1=4的单调递增数列,且满足an+1^2+an^2+16=8(an+1+an)+2an+1an,求ann+1均为a的下标
已知数列an满足条件a1=-2 a(n+1)=2an/(1-an) 则an=
已知数列an满足a1=1,a(n+1)=an/(3an+2),则an=?