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

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

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