正数数列an的前n项和为sn,且对任意n=N+都有2sn+nan=6n+3,求a1,a2,a3;求an的通项公式;令bn=(an-2)(n+1),bn的前n项和为Tn,求证Tn扫码下载作业帮搜索答疑一搜即得

问题描述:

正数数列an的前n项和为sn,且对任意n=N+都有2sn+nan=6n+3,求a1,a2,a3;求an的通项公式;
令bn=(an-2)(n+1),bn的前n项和为Tn,求证Tn

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搜索答疑一搜即得

(1)
2Sn+nan=6n+3
n=1
3a1=9
a1=3
n=2

(1)2Sn+nan=6n+3n=13a1=9a1=3n=22(a1+a2)+2a2=154a2=9a2=9/4n=32(a1+a2+a3)+3a3=215a3=21/2a3=21/10 2Sn+nan=6n+3 (1)2S(n-1)+(n-1)a(n-1)=6(n-1)+3 (2)(1)-(2)2an+nan-(n-1)a(n-1)=6(2+n)[an -2] ...