A1=1Sn=[(1+An)/2]^2S(n-1) =[(1+A(n-1))/2]^2然后减一下得到An = [An^2 +2An - A(n-1)^2 - 2A(n-1)]/4化简得到(An - 1)^2 = (A(n-1) +1)^2An = A(n-1) +2,故a1=1,d=2An = 2n-1bn=a(2n)Tn=(1+4n-1)*2n/2=(2n)^2=4n^2...
已知等差数列{an}的公差为d,前n项和为Sn,并且Sn=((1+an)/2)².求a1,d
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