Question

已知数列{a_n}前n项和为S_n,此无穷数列对于不小于2的正整数n,满足1-S_n=a_n-1-a_n.

(1)求a₁,a₂,a₃;

(2)证明{a_n}为等比数列;

(3)设,求(b₁+b₂+…+b_n)的值.

Answer 1

(1)解析:∵S₂=a₁+a₂,?

∴1-(a₁+a₂)=a₁-a₂,?

解得a₁=.?

∵S₃=a₁+a₂+a₃,?

同理解得a₂=,a₃=.?

(2)证明:当n≥2时,1-S_n=a_n-1-a_n.      ①?

1-S_n+1=a_n-a_n+1 .                          ②?

①-②得S_n+1-S_n=a_n-1-2a_n+a_n+1 ,?

∴a_n+1 =a_n-1 -2a_n+a_n+1 .?

∴,?

即.?

∴{a_n}为等比数列.?

(3)解析:∵b_n=?

,?

∴(b₁+b₂+…+b_n)?

=

=.