设数列{an}的前n项和为Sn,a1=1,且3a(n+1)=Sn,(1)求a2,a3,a4及an (2)求a2+a4+···+a2n

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设数列{an}的前n项和为Sn,a1=1,且3a(n+1)=Sn,(1)求a2,a3,a4及an (2)求a2+a4+···+a2n
设数列{an}的前n项和为Sn,a1=1,且3a(n+1)=Sn,(1)求a2,a3,a4及an (2)求a2+a4+···+a2n

设数列{an}的前n项和为Sn,a1=1,且3a(n+1)=Sn,(1)求a2,a3,a4及an (2)求a2+a4+···+a2n
1、通过3a(n+1)=Sn,
n=1时,得到3*a2=a1,即a2=1/3;( * 表示乘号)
n≥2时,Sn=3*a(n+1),S(n-1)=3*an,
两式相减得到an=3*a(n+1)-3*an,即a(n+1)=(4/3)*an,
即a3=(4/3)*a2=4/9,a4=(4/3)*a3=16/27,
an=(4/3)*a(n-1)=(4/3)^2*a(n-2)=...=(4/3)^(n-2)*a2=(4/3)^(n-2)*(1/3)
=4^(n-2)/3^(n-1),( ^ 表示N次方)
（实际上这是一个从第三项开始公比为4/3的等比数列）
2、记Bn=a2+a4+···+a2n,
根据Sn=3*a(n+1),an=(4/3)*a(n-1),得到Sn=3*a(n+1)=3*(4/3)*a(n)=4*an,
S(2n+1)=a1+a2+a3+a4+a5+...+a(2n-1)+a2n+a(2n+1)
=1+a2+4/3*a2+a4+4/3*a4+...+a2n+4/3*a2n
=1+(7/3)*(a2+a4+...+a2n)=1+(7/3)*Bn
=4*a(2n+1)=4*4^(2n-1)/3^2n-=4^2n/3^2n
Bn=(4^2n/3^2n-1)*3/7
=4^2n/[7*3^(2n-1)]-3/7

S(n+1)=Sn+a(n+1)=4a(n+1)=3a(n+2),即an/a(n-1)=4/3,an=(4/3)^(n-1)
(1):a2=4/3,a3=16/9,a4=64/27
(2)令bn=a(2n)=(4/3)^(2n-1)=(3/4)*(16/9)^n,
Bn=(12/7)*((16/9)^n-1)

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