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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
人气:414 ℃ 时间:2019-08-19 16:40:50
解答
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+...
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