问题
解答题
设等差数列{an}的前n项和为Sn,已知a2=3,S5=25
(1)求数列{an}的通项an
(2)设数列{bn}满足bn=2an求数列{bn}的前n项和Tn.
答案
(1)设等差数列首项为a1公差为d
由a2=a1+d=3 s5=5a1+10d=25
解得a1=1 d=2
an=1+2(n-1)=2n-1
(2)bn=2an=22n-1
Tn=bn+bn+bn+…+bn
=21+23+25+…+22n-1=21(4n-1) 4-1
=2(4n-1) 3