已知各项均为正的数列为等比数列,;为等差数列的前n项和,. (1

已知各项均为正的数列为等比数列,为等差数列的前n项和,.

1 的通项公式;

2 ,求.

答案

解:(1 {an}的公比为q,由a5=a1q4q=4所以an=4n-1.

{ bn }的公差为d,由5S5=2 S85(5 b1+10d)=2(8 b1+28d),

所以bn=b1+(n-1)d=3n-1

2 Tn=1·2+4·5+42·8+…+4n-1(3n-1),   4Tn=4·2+42·5+43·8+…+4n(3n-1),

-①得:3Tn=-2-3(4+42+…+4n-1)+4n(3n-1)= -2+4(1-4n-1)+4n(3n-1)=2+(3n-2)·4n

Tn=n-4n+

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