已知正项等差数列已知正项等差数列an的前n项和为sn,若s3=12,2a1,a2,a3+1成等比数列.

问题描述:

已知正项等差数列已知正项等差数列an的前n项和为sn,若s3=12,2a1,a2,a3+1成等比数列.
(1)求{an}的通项公式
(2)记bn=2^an+3前n项和Tn,求Tn.

(1)∵S3=12即 a1+a2+a3=3a2=12∴ a2=4又2a1,a2,a3+1成等比数列∴(a2)²=2a1*(a3+1)=2(a2-d)(a2+d+1)∴ 4=2*(4-d)*(5+d)∴d²-d-12=0∴ d=3或d=-4 (舍)∴ an=3n+1(2)bn=2^(3n+4)=16*8^nTn=16*[8-8^(n+1)]...