设等差数列{An}的前n项和为Sn,且S4=4S2,A2n=2An+...

(Ⅰ)设等差数列{an}的首项为a1,公差为d,由S4=4S2,a2n=2an+1得:4a1+6d=

(1)设等差数列{an}的首项为a1,公差为d,由a2n=2an+1,取n=1,得a2=2a1+1,

(1)∵等差数列{an}的前n项和为Sn.且S4=4S2,a2n=2an+1,∴4a1+4?32d=

An为等差数列,则A2n=A1+(2n-1)d;An=A1+(n-1)d 又因为A2n=2An+1

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