已知数{An}满足An+1=3An+3ˆ(n+1)-1(n∈N+),且A4=365,求数列{An}通项An
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已知数{An}满足An+1=3An+3ˆ(n+1)-1(n∈N+),且A4=365,求数列{An}通项An
如题,脚码打不出来,
如题,脚码打不出来,
A(n+1)=3An+3*(n+1)-1 ======>>>
==>>>> A(n+1)-(1/2)×3^(n+2)-(1/2)=3An-(3/2)×3^(n+1)-(3/2)
==>>>> A(n+1)-(1/2)×3^(n+2)-(1/2)=3[An-(1/2)×3^(n+1)-(1/2)]
===>>>> [A(n+1)-(1/2)×3^(n+2)-(1/2)]/[An-(1/2)×3*(n+1)-(1/2)]=3=常数
则数列{An-(1/2)×3^(n+1)-(1/2)}是以A1-(1/2)×3²-(1/2)为首项、以q=3为公比的等比数列.这个等比数列的第四项是A4-(1/2)×3^5-(1/2)=243=243=3^5
An-(1/2)×3^(n+1)-(1/2)=[3^5]×3^(n-4)=3^(n+1)
An=(1/2)×[3^(n+2)+1]
==>>>> A(n+1)-(1/2)×3^(n+2)-(1/2)=3An-(3/2)×3^(n+1)-(3/2)
==>>>> A(n+1)-(1/2)×3^(n+2)-(1/2)=3[An-(1/2)×3^(n+1)-(1/2)]
===>>>> [A(n+1)-(1/2)×3^(n+2)-(1/2)]/[An-(1/2)×3*(n+1)-(1/2)]=3=常数
则数列{An-(1/2)×3^(n+1)-(1/2)}是以A1-(1/2)×3²-(1/2)为首项、以q=3为公比的等比数列.这个等比数列的第四项是A4-(1/2)×3^5-(1/2)=243=243=3^5
An-(1/2)×3^(n+1)-(1/2)=[3^5]×3^(n-4)=3^(n+1)
An=(1/2)×[3^(n+2)+1]
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