Question

等差数列{a_n}的前n项和为S_n，且S₄=5a₂,a₃=3,令b_n=1S_n，n∈N^*.

(1)求数列{b_n}的通项公式；

(2)求T_n=b₁+b₂+…+b_n.

Answer 1

解:（1）方法一：设等差数列{a_n}的公差为d，依题意，得

?

解得a₁=1,d=1.                                                                                                        ?

∴S_n=.∴b_n=.                                                                                 ?

方法二:设等差数列{a_n}的公差为d.?

由S₄=a₁+a₂+a₃+a₄=2(a₂+a₃),?

又∵S₄=5a₂∴2(a₂+a₃)=5a₂.                                                                                      ?

∵a₃=3,?

解得a₂=2.                                                                                                               ?

∴a₁=1,d=1.                                                                                                            ?

∴a_n=n.?

∴S_n=n+·1=.∴b_n=.                                                           ?

（2）T_n=2［(1-)+(-)+…+(-)］                                                        ?

=2(1-)=.