Question

在數(shù)列{a_n}中,已知a_n+1a_n=2a_n-a_n+1,且a₁=2(n∈N^*),

(1)求證:數(shù)列{-1}是等比數(shù)列;

(2)設(shè)b_n=a_n²-a_n,且S_n為{b_n}的前n項(xiàng)和,試證:2≤S_n＜3.

Answer 1

證明:(1)由a_n+1a_n=2a_n-a_n+1且a₁=2≠0,得a_n＞1≠0(n∈N^*).

再由等式兩邊同除以a_n+1a_n,得-1=(-1).

由a₁=2得-1=.所以數(shù)列{-1}是首項(xiàng)為,公比為的等比數(shù)列.

(2)方法一:由(1)知-1=()^n-1=-()ⁿ,即a_n=.

故a_n²-a_n=a_n(a_n-1)==b_n.

而S_n+1-S_n=b_n+1=＞0,故S_n是關(guān)于n的遞增數(shù)列.

故S_n≥S₁=b₁=a₁²-a₁=2²-2=2.8分當(dāng)k≥2時(shí),b_k=a_k(a_k-1)=＜

==.

故S_n＜+=3＜3.

綜上,有2≤S_n＜3.

方法二:∵b_n=a_n²-a_n=a_n(a_n-1)(而a_n+1=,故1＜a_n≤2)≤2(a_n-1)=2(-1)=.

∴S_n=b₁+b₂+b₃+b₄+…+b_n≤2++++c+…+

＜2++++=2++++＜2++++

=2++++=2+＜3.

綜上,有2≤S_n＜3.