an/n=2 an=2n n=1时，a1=2×1=2，同样满足表达式 数列{an}的通项公式为an=2n (2)4/[an(an+2)]=4/[2n×(2n+2)]=1/[n(n+1)]=1/n -1/(n+1)Tn=1-½+½-⅓+...+1/n -1/(n+1)=1- 1/(n+1)1/(n+1)>0，1- 1/(n+1)<1 随n增大，n+1...
所以，2Sn+1=Sn+4① 当n≥2时，2Sn=Sn-1+4② ①-②，得an+1＝ 1 2 an（n≥2），又a2＝ 1 2 a1，所以数列{an}是首项为2，公比为 1 2 的等比数列．所以{an}的通项公式为an＝(1 2 )n?2（n∈N*）．（2）由（1），得Sn＝4(1?1 2n )，由 Sn?m Sn+1?m ＝ 1 am+1...
(12分)设数列{an}的前n项和为Sn.已知a1=1,2Sn=nan+1﹣,n∈N*.(Ⅰ) 求数列{an}的通项公式;(Ⅱ) 证明:对一切正整数n,有.
2an-a1 = S1.Sn 2[Sn-S(n-1)]-1 = Sn Sn +1 = 2[S(n-1)+ 1]{Sn +1 } 是等比数列,q=2 Sn +1 = 2^(n-1).(S1 +1 )= 2^n Sn = -1+2^n an = Sn -S(n-1)= 2^(n-1)bn = nan = n.2^(n-1)let S = 1.2^0+ 2.2^1+...+n.2^(n-1)(1)...
解答(I)解:∵2Sn=(n+1)an, ∴当n≥2时,2Sn-1=nan-1,可得2an=(n+1)an-nan-1, ∴annann=an−1n−1an−1n−1. ∴annann=a11a11, ∴an=2n. (II)证明:4an(an+2)4an(an+2)=42n(2n+2)42n(2n+2)=1n−1n+11n−1n+1. ...
2s(n) + 2 = a(n+1) = s(n+1) - s(n),s(n+1) = 3s(n) + 2,s(n+1) + 1 = 3[s(n)+1],{s(n)+1}是首项为s(1)+1=a(1)+1=3,公比为3的等比数列。s(n)+1 = 3*3^(n-1) =3^n.s(n)=3^n - 1.a(n+1) = 2s(n) + 2 = 2*3^n - 2 + 2 ...
【题目】已知数列{an}的前n项和为Sn,a1=2且an+1=2Sn+2(n∈N*).(1)求Sn2)设bn=,求数列{bn}的前n项和Tnn+1
2 Sn S(n-1),两边同除以Sn S(n-1)可得：1/ Sn -1/ S(n-1)=2,所以数列{1/ Sn }是等差数列,首项为1,公差为2,1/ Sn=1+2(n-1)=2n-1,Sn=1/(2n-1)∴n=1时,a1=1,n≥2时,an= Sn- S(n-1)= 1/(2n-1)- 1/(2n-3)=-2/[(2n-1) (2n-3)].希望能帮到您,
∴当n≥2时,2Sn-1=nan-1,可得2an=(n+1)an-nan-1,∴ an n= an-1 n-1.∴ an n= a1 1,∴an=2n.(II)证明: 4 an(an+2)= 4 2n(2n+2)= 1 n- 1 n+1.∴Tn= (1- 1 2)+ ( 1 2- 1 3)+…+ ( 1 n- 1 n+1)=1- 1 n+1.∴ 1 2=T1≤Tn<1,∴ 1 2≤Tn<1. 解析...
设数列{an}的前n项和为Sn,已知a1=1,2Sn=(n+1)an,n∈N*.(1)求数列{an}的通项公式;(2)令bn=,数列{bn}的前n项和为Tn,试比较Tn与