1)前2n项和为(2n)^3=8n^3, 前n个奇数项和=前2n项和-前n项偶数和=8n^3-(n^2)(4n+3)=n^2(4n-3) 之前看错题目了,选B吧
2)由题n*S n+1=2n(n+1)+(n+1)Sn,两边同时除以n(n+1) 得S n+1 /(n+1)=2+ Sn/n令bn=Sn/n b n+1=2+bn则bn是等差数列,公差为2b1=S1/1=15 Sn/n=15+2(n-1)=2n+13Sn=2n^2+13nan=Sn- S n-1=2(n^2-(n-1)^2)+13(n-(n-1))=2(2n-1)+13=4n+11PQ的其中一个方向向量为+ -[(n+2,a n+2)-(n,an)]=+ -(2,a n+2-an)=+ -(2,8)易知(-1/2,-2)与-(2,8)同向,故选D
3)bn=|(an+2)/(an-1)|=|(an+1+1)/(an+1-2)|=|(2/a n+1 +1)/(2/an+1 -2)|=|(a n+1 +2)/(-2a n+1 -2)|=1/2*|(a n+1 +2)/(a n+1 -1)|=1/2b n+1
即{bn}是公比为1/2的等比数列 b1=|(a1+2)/(a1-1)|=4 bn=1/2*4^(n-1)
4)由an=2Sn^2/(2sn-1),有(2sn-1)an=2Sn^2由于Sn=S n-1 +an 代入(2sn-1)an=2Sn^2 得(2s n-1 +an)an=2(S n-1 +an)^2 整理一下有 -an=2S n-1 *(Sn-1+an)=2S n-1 *Sn 即-an=S n-1 -Sn=2S n-1 *Sn 两边同除以Sn*S n-1 得1/Sn-1/S n-1=2 {1/Sn}成等差数列 首项为1/S1=1 公差为2 1/Sn=1+2(n-1)=2n-1 Sn=1/(2n-1) an=Sn-S n-1=1/(2n-1)-1/(2n-3)=-2/[(2n-1)(2n-3)]