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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=

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

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

设数列{an}的公差为d(d≠0),首项为a1,由已知得:(3a1+3d)2=9(2a1+d)4a1

d=a4-a3=(s4-s3)-(s3-s2)=s4+s2=5s2=5(a1+a2)=10a1+5d

设函数an=a1*q^(n-1),又4s1,3s2,2s3成等差数列,那么有 2*3s2=4s1+

(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,

(Ⅰ)设等差数列{an}的公差为d,则4a1+4×32d=4(2a1+d)a1+(2n-1)d=2a

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