已知数列{an}满足,an=3an-1+3^2-1,且a3=95是否存在一个实数t,使得bn=1\3^n(an+t)且{
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已知数列{an}满足,an=3an-1+3^2-1,且a3=95是否存在一个实数t,使得bn=1\3^n(an+t)且{bn}为等差数列
如果原题中“an=3an-1+3^2-1”为an=3an-1+3^n-1,
则由an=3an-1+3^n-1
两边同除以3^n,化为
an/3^n=an-1/3^(n-1)+1-3^(-n)
则a2/3^2=a1/3^(2-1)+1-3^(-2)
a3/3^3=a2-1/3^(3-1)+1-3^(-3)
……
an/3^n=an-1/3^(n-1)+1-3^(-n)
上述(n-1)个式子累加化为:
an/3^n=a1/3^(2-1)+(n-1)-[3^(-2)+3^(-3)+……+3^(-n)]
=a1/3+(n-1)-[1/3-3^(-n)]
=a1/3+n-4/3+3^(-n)
则n>=2时,an=3^n*[a1/3+n-4/3+3^(-n)]
则a3=27*[a1/3+3-4/3+1/27]]=95,解得a1=79/9
an=3^n*[79/27+n-4/3+3^(-n)]
则bn=1\3^n(an+t)=(an+t)/3^n={3^n*[79/27+n-4/3+3^(-n)]+t}/3^n
bn要为等差数列,只需3^n*3^(-n)]+t=0,则t=-1
则由an=3an-1+3^n-1
两边同除以3^n,化为
an/3^n=an-1/3^(n-1)+1-3^(-n)
则a2/3^2=a1/3^(2-1)+1-3^(-2)
a3/3^3=a2-1/3^(3-1)+1-3^(-3)
……
an/3^n=an-1/3^(n-1)+1-3^(-n)
上述(n-1)个式子累加化为:
an/3^n=a1/3^(2-1)+(n-1)-[3^(-2)+3^(-3)+……+3^(-n)]
=a1/3+(n-1)-[1/3-3^(-n)]
=a1/3+n-4/3+3^(-n)
则n>=2时,an=3^n*[a1/3+n-4/3+3^(-n)]
则a3=27*[a1/3+3-4/3+1/27]]=95,解得a1=79/9
an=3^n*[79/27+n-4/3+3^(-n)]
则bn=1\3^n(an+t)=(an+t)/3^n={3^n*[79/27+n-4/3+3^(-n)]+t}/3^n
bn要为等差数列,只需3^n*3^(-n)]+t=0,则t=-1
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