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,根号S2.根号S3成等差数列.求a

a2=a1+d=a1+2 a3=a1+4 s1=a1 s2=a1+a2=2a1+2 s3=a1+a2+a3=3a1+6 根号S1+根号S3=2倍根号S2 根号a1+根号(3a1+6)=2倍根号(2a1+2) a1+3a1+6+2倍根号[a1(3a1+6)]=8a1+8 2a1+1=根号[a1(3a1+6)] 4a1平方+4a1+1=3a1平方+6a1 a1平方-2a1+1=0 a1=1 所以an=a1+（n-1)d=2n-1
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