已知数列{An}前n项和为Sn,A1=10,A(n+1)=9Sn+10.1,求证{lgAn}是等差数列.
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已知数列{An}前n项和为Sn,A1=10,A(n+1)=9Sn+10.1,求证{lgAn}是等差数列.
2,若Tn是数列{3/(lgAn)(lgA(n+1))的前n项和,求Tn
注:
2,若Tn是数列{3/(lgAn)(lgA(n+1))的前n项和,求Tn
注:
An=9Sn-1+10,An+1-An=9Sn-9Sn-1=9An,所以,An=10An-1,A1=10,所以,An=10^n,lgAn=n.
2问,设Cn=3/(lgAn)(lgA(n+1),则Cn=3/n(n+1),Tn=3(1/1*2+1/2*3+1/3*4+.+1/n(n+1)),Tn=3(2-1/1*2+3-2/2*3+4-3/3*4+.+n+1-n/n(n+1))=3(1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ...+ 1/n - 1/(n+1))=3(1- 1/(n+1))=3n / (n+1)
2问,设Cn=3/(lgAn)(lgA(n+1),则Cn=3/n(n+1),Tn=3(1/1*2+1/2*3+1/3*4+.+1/n(n+1)),Tn=3(2-1/1*2+3-2/2*3+4-3/3*4+.+n+1-n/n(n+1))=3(1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ...+ 1/n - 1/(n+1))=3(1- 1/(n+1))=3n / (n+1)
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