Question

在數(shù)列{a_n}中,已知a₁=2,a_n+1=3a_n+λn-1,n∈N^*,λ為常數(shù).

(1)若數(shù)列{a_n+n}是等比數(shù)列,求實數(shù)λ的值;

(2)在(1)的條件下,求數(shù)列{a_n}的前n項和S_n.

Answer 1

解:(1)由a₁=2且a_n+1=3a_n+λn-1得a₂=3a₁+λ-1=5+λ,a₃=3a₂+2λ-1=5λ+14.

∵數(shù)列{a_n+n}是等比數(shù)列,∴(a₂+2)²=(a₁+1)(a₃+3).2分∴(λ+7)²=3(5λ+17),

整理得λ²-λ-2=0,解得λ=2或λ=-1.

當λ=2時,由a_n+1=3a_n+2n-1得a_n+1+n+1=3(a_n+n),

∴=3.又a₁+1=3,

∴數(shù)列{a_n+n}是首項為3,且公比為3的等比數(shù)列.

〔或當λ=2時,由a_n+1=3a_n+2n-1得===3〕

當λ=-1時,≠常數(shù).

∴當數(shù)列{a_n+n}是等比數(shù)列時,λ=2.

(2)由(1)可知a_n+n=3×3^n-1=3ⁿ,∴a_n=3ⁿ-n.

∴數(shù)列{a_n}的前n項和S_n=(3+3²+…+3ⁿ)-(1+2+…+n)=.