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
2个回答
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.
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