问题描述: 设数列{an}是首项为a1(a1>0),公差为2的等差数列,其前n项和为Sn,且根号S1,S2,S3成等差数列.求数列{an}的通项公式 1个回答 分类:数学 2014-10-03 问题解答: 我来补答 Sn=a1n+n(n-1)2/2=a1n+n(n-1)2根号S2=根号S1+根号S32根号(2a1+2)=根号a1+根号(3a1+6)4(2a1+2)=a1+3a1+6+2根号a1(3a1+6)8a1+8=4a1+6+2根号a1(3a1+6)4a1+2=2根号a1(3a1+6)2a1+1=根号a1(3a1+6)4a1^2+4a1+1=3a1^2+6a1a1^2-2a1+1=0(a1-1)^2=0a1=1an=a1+(n-1)d=1+(n-1)2=2n-1 展开全文阅读