Question

已知數(shù)列{a_n}滿足a₁=a,a_n+1=(n∈N^*).

(1)判斷數(shù)列{}是否為等比數(shù)列？若不是,請(qǐng)說(shuō)明理由;若是,試求出通項(xiàng)a_n;

(2)如果a=1時(shí),數(shù)列{a_n}的前n項(xiàng)和為S_n,試求出S_n,并證明當(dāng)n≥3時(shí),有++…+＜.

Answer 1

解:(1)∵a_n+1+2=+2=,

∴=2·.令b_n=,則b_n+1=2b_n.

∵b₁=,∴當(dāng)a=-2時(shí),b₁=0,則b_n=0.∵數(shù)列{0}不是等比數(shù)列,

∴當(dāng)a=-2時(shí),數(shù)列{}不是等比數(shù)列.

當(dāng)a≠-2時(shí),b₁≠0,則數(shù)列{}是等比數(shù)列,且公比為2.

∴b_n=b₁·2^n-1,即=·2^n-1.解得a_n=·2^n-1-2.

(2)由(1)知,當(dāng)a=1時(shí),a_n=(2n+1)·2^n-1-2,S_n=3+5·2+7·2²+…+(2n+1)·2^n-1-2n.

令T_n=3+5·2+7·2²+…+(2n+1)·2^n-1,①

則2T_n=3·2+5·2²+…+(2n-1)·2^n-1+(2n+1)·2ⁿ,②

由①-②,得-T_n=3+2(2+2²+…+2^n-1)-(2n+1)·2ⁿ

=3+2·-(2n+1)·2ⁿ=(1-2n)·2ⁿ-1,

∴T_n=(2n-1)·2ⁿ+1,9分則S_n=T_n-2n=(2n-1)(2ⁿ-1).∵2ⁿ=++…++Cⁿ_n,∴當(dāng)n≥3時(shí),2ⁿ≥+++Cⁿ_n=2(n+1),

則2ⁿ-1≥2n+1.12分∴S_n≥(2n-1)(2n+1),則≤=().

因此,++…+≤［()+()+…+()］=()＜.