已知数列an的前n项和为sn,a1=2,nan+1=sn+n(n+1),设bn=sn/2n,bn小于等于t,
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/17 17:54:35
![已知数列an的前n项和为sn,a1=2,nan+1=sn+n(n+1),设bn=sn/2n,bn小于等于t,](/uploads/image/z/644170-58-0.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BAsn%2Ca1%3D2%2Cnan%2B1%3Dsn%2Bn%EF%BC%88n%2B1%EF%BC%89%2C%E8%AE%BEbn%3Dsn%2F2n%2Cbn%E5%B0%8F%E4%BA%8E%E7%AD%89%E4%BA%8Et%2C)
xPN@~6lٞ 6'x3&hV pRD=Pa[S_-I67}fEA=}q5[P<K0G
6m:YhOiCQSnto4_3CCTn\l"rDHz9SAۖQtwM!82Q^TS8گH S7P=/+GaF^=ρ)SM,x1dMZxSC
XPoƪߖw< [M/0҅μRjZVZ
已知数列an的前n项和为sn,a1=2,nan+1=sn+n(n+1),设bn=sn/2n,bn小于等于t,
已知数列an的前n项和为sn,a1=2,nan+1=sn+n(n+1),设bn=sn/2n,bn小于等于t,
已知数列an的前n项和为sn,a1=2,nan+1=sn+n(n+1),设bn=sn/2n,bn小于等于t,
na(n+1)=n[S(n+1)-Sn]=Sn+n(n+1),即nS(n+1)=(n+1)Sn+n(n+1),两边除以n(n+1),得:[S(n+1)]/(n+1)-[Sn]/n=1=常数,则{(Sn)/n}是以(S1)/1=a1=2为首项、以d=1为公差的等差数列,得:(Sn)/n=n+1,所以Sn=n(n+1),bn=(Sn)/(2n)=(n+1)/2,……
问题呢,求an,bn?
求什么?
通项、、
数列An的前n项和为Sn,已知A1=1,An+1=Sn*(n+2)/n,证明数列Sn/n是等比数列
已知数列{an}的前n项和为Sn,若a1=1/2,Sn=n^2an-n(n-1)求Sn,an
数列:已知数列{an}前 n项和为Sn,且a1=2,4Sn=ana(n+1).求数列{an}的通项公式.
已知Sn为数列的前n项和,a1=2,2Sn=(n+1)an+n-1,求数列an的通项公式
已知数列{an}的前n项和为Sn,又a1=2,nAn+1=sn+n(n+1),求数列{an}的通项公式
已知数列 an前n项和为Sn,a1=1,Sn=2a(n+1),求Sn
已知数列《an>的前n项和为sn,a1=2,na=sn,求s2011
已知数列{an}a1=2前n项和为Sn 且满足Sn Sn-1=3an 求数列{an}的通项公式an已知数列{an}a1=2前n项和为Sn 且满足Sn +Sn-1=3an 求数列{an}的通项公式an
已知数列{an}的前n项和为Sn,a1=1/2,且Sn=n^2An-n(n-1),求an
已知数列{an}中,a1=2,前n 项和为Sn,若Sn=n^2*an,
已知数列的前N项和为SN,A1=2,2sn的平方=2ansn-an(n≥2)求an和sn
已知数列An中,其前n项和为Sn,A1=1,且An+1=2Sn,求An的通项公式和Sn
已知数列{an}的前n项和为Sn,且满足an+2Sn*Sn-1=0,a1=1/2.求证:{1/Sn}是等差数列
数列{an}的前n项和为Sn,已知a1+2,Sn+1=Sn-2nSn+1Sn,求an紧急紧急!求救中!sos
已知数列{an} 的前n项和为sn,且an=sn *s(n-1)a1=2/9 求证:{1/sn}为等差
已知数列{an}的前n项和记为sn,且a1=2,an+1=sn+2.求数列an的通项公式.
数列:已知数列[An]前n项和为Sn a1=1 An+1=2Sn 求【An] 求【n-An]前n项和Sn数列:已知数列[an]前n项和为Sn,a1=1 ,a[n+1]=2Sn,求[an]通项,求[n-an]前n项和Sn.注:a[n+1]指a 的下标为n+1而不是以n为下标的a加上1.
已知数列{an}的前n项和为Sn,a1=1/2,Sn=n的平方*an,求a1,a2,