设数列的{an}前n项和为Sn 且满足2a（n）= 3Sn-5/2S(n-1)-2(n>=2) a(1)=2.(1) 求

来源：学生作业帮编辑：拍题作业网作业帮分类：数学作业时间：2024/04/25 17:41:00

设数列的{an}前n项和为Sn 且满足2a（n）= 3Sn-5/2S(n-1)-2(n>=2) a(1)=2.(1) 求数列{an}的通项公式(2)证明:1/2(log(2)Sn+log(2)S(n＋2））＜log(2)S(n+1)

2a（n）= 3Sn-5/2S(n-1)-2
即2[Sn-S(n-1)]=3Sn-(5/2)S(n-1)-2
Sn=(1/2)S(n-1)+2
Sn-4=(1/2)[S(n-1)-4]
所以{Sn-4}是公比为1/2的等比数列
首项=S1-4=2-4=-2
所以Sn-4=(-2)*(1/2)^(n-1)=
故Sn=-2/2^(n-1)+4
S(n-1)=-2/2^(n-2)+4
所以通项公式an=Sn-S(n-1)=1/2^(n-2)
log2 Sn=log2 [4-1/2^(n-2)]=2-log2 (1-1/2^n)
log2 S(n+2)=log2 [4-1/2^n]=2-log2 [1-1/2^(n+2)]
log2 S(n+1)=log2 [4-1/2^(n-1)]=2-log2 [1-1/2^(n+1)]
log2 Sn+log2 Sn=4-log2 (1-1/2^n)[1-1/2^(n+2)]
2log2 S(n+1)=4-2log2 [1-1/2^(n+1)]=4-2log2 [1-1/2^(n+1)]^2
要使不等式成立,只需 [1-1/2^(n+1)]^2>(1-1/2^n)[1-1/2^(n+2)]
即1-1/2^n+1/2^(2n+2)>1-1/2^n-1/2^(n+2)+1/2^(2n+2)
亦即-1/2^n>-1/2^n-1/2^(n+2)
显然成立
得证.

已知数列an的首项a1=5,前n项和为Sn,且S(n+1)=2Sn+n+5（n∈N*）,求数列｛an｝的前n项和Sn,设已知数列{an}的前n项和为Sn,且a1=2,3Sn=5an-A(n-1)+3S(n-1)(n≥2,n属于N*)设bn= 设数列{an}的前n项和Sn,已知首项a1=3,且S(n+1)+Sn=2a(n+1),求此数列的通项公式和前n项和Sn 设数列an的前n项和为sn,满足2Sn=a（n+1）-2^(n+1)+1,n∈N,且a1,a2+5,a3成等差数列高中数列难题.设数列{an}的前n项和为sn,满足2sn=a(n+1)-2^(n+1)+1,n属于n*.且a1,a2+5 设数列{an}的前n项和为Sn,并且满足2Sn=an²+n,an>0.(1)求a1,a2,a3.(2)猜想{a 设数列an的前n项和为Sn.已知首项a1等于3,且S(n+1)+Sn=2a(n+1)求通项公式以及前n项和sn 设数列an的前n项和为Sn,已知a1=1,(2Sn)/n=a(n+1)-1/3n^2-n-2/3 设数列an的前n项和Sn.已知首项a1=3,S(n+1)+Sn=2a(n+1),试求此数列的通向同事an和前n项和Sn 设数列{an}的前n项和为sn,满足2sn=a(n+1)-2^(n+1)+1,n属于n*.且a1,a2+5,a3成等差数已知数列an前n项和为Sn,且满足4（n+1)(Sn+1)=(n+2)^2an(n属于正整数）求an 已知数列{an}的首项a1=3,前n项和为Sn,且S（n+1）=3Sn+2n(n∈N)