数学…求通项公式

S2n/Sn=(4n+2)/(n+1)=n*(4n+2)/[n*(n+1)]=2n*(2n+1)/[n*(n+1)]

S2n = 2n*(2n+1)*k, Sn=n*(n+1)*k

S1=1*(1+1)*k=A1=1, k=1/2, 即Sn=n*(n+1)/2

An=Sn-S(n-1)=n*(n+1)/2-n*(n-1)/2=n

即An＝n

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Sn=n+n(n-1)d/2
S2n=2n+(2n)(2n-1)d/2
S2n/Sn=[2n+(2n)(2n-1)d/2]/[n+n(n-1)d/2]=(4n+2)/(n+1)
(4+2(2n-1)d)/(2+(n-1)d)=(4n+2)/(n+1)
4d=4,4-2d=2,d=1,2-d=1
d=1
an=n

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