已知函数f(x)=x/(3x+1),数列{an}满足a1=1,an+1=fan)(n∈N*),若数列{bn}的前n项和S
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已知函数f(x)=x/(3x+1),数列{an}满足a1=1,an+1=fan)(n∈N*),若数列{bn}的前n项和Sn=2^n-1,记Tn=b1/a1+b2/a2+.+bn/an,求Tn
an+1 = an / (3an + 1)
1/an+1 = 3 + 1/an
令cn = 1/an
则cn+1 = 3 + cn
c1 = 1/a1 = 1
故cn = 3n-2
an = 1/(3n-2)
Sn = 2^n - 1
Sn-1 = 2^(n-1) - 1
bn = Sn - Sn-1 = 2^n - 2^(n-1) = 2^(n-1)
Tn = 1/1 + ...+ 2^(n-1)*(3n-2)
=1 + (2*3*2 - 2²) + ...+ [3*n*2^(n-1) - 2^n]
=3*[1*(2^n-1)/(2-1) + 2*[2^(n-1)-1]/(2-1) + ...+ 2^(n-1)*[2^(n-n+1)-1]/(2-1)] - 2*(2^n-1)/(2-1)
=3*[n*2^n - 1*(2^n-1)/(2-1)] - 2^(n+1) + 2
=3n*2^n - 3*2^n + 3 - 2*2^n + 2
=(3n-5)2^n + 5
得到Tn = (3n-5)2^n + 5
1/an+1 = 3 + 1/an
令cn = 1/an
则cn+1 = 3 + cn
c1 = 1/a1 = 1
故cn = 3n-2
an = 1/(3n-2)
Sn = 2^n - 1
Sn-1 = 2^(n-1) - 1
bn = Sn - Sn-1 = 2^n - 2^(n-1) = 2^(n-1)
Tn = 1/1 + ...+ 2^(n-1)*(3n-2)
=1 + (2*3*2 - 2²) + ...+ [3*n*2^(n-1) - 2^n]
=3*[1*(2^n-1)/(2-1) + 2*[2^(n-1)-1]/(2-1) + ...+ 2^(n-1)*[2^(n-n+1)-1]/(2-1)] - 2*(2^n-1)/(2-1)
=3*[n*2^n - 1*(2^n-1)/(2-1)] - 2^(n+1) + 2
=3n*2^n - 3*2^n + 3 - 2*2^n + 2
=(3n-5)2^n + 5
得到Tn = (3n-5)2^n + 5
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