已知数列(an)满足a1=1.an+1=3an+2n-1,求an
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已知数列(an)满足a1=1.an+1=3an+2n-1,求an
已知数列(an)满足a1=1.an+1=3an+2n-1,求an
已知数列(an)满足a1=1.an+1=3an+2n-1,求an
a(n+1)=3an+(2n-1),a(n+1)+(n+1)=3an+3n=3(an+n),则:[a(n+1)+(n+1)]/[an+n]=3=常数,则数列{an+n}是以a1+1=2为首项、以q=3为公比的等比数列,得:an+n=2×3^(n-1),即:an=2×3^(n-1)-n
an+1=3an+2n-1,an+1+(n+1)=3an+3n=3(an+n)
[an+1+(n+1)]/[an+n]=3
故:数列{an+n}是以a1+1=2为首项、以q=3为公比的等比数列,
得:an+n=2×3^(n-1)
所以:an=2×3^(n-1)-n
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