Question

在數(shù)列{a_n}中,已知a₁=-1,且a_n+1=2a_n+3n-4(n∈N^*).

(1)求證:數(shù)列{a_n+1-a_n+3}是等比數(shù)列;

(2)求數(shù)列{a_n}的通項(xiàng)公式;

(3)求和:S_n=|a₁|+|a₂|+|a₃|+…+|a_n|(n∈N^*).

Answer 1

答案：(1)證明:令b_n=a_n+1-a_n+3b_n+1=a_n+2-a_n+1+3=2a_n+1+3(n+1)-4-2a_n-3n+4+3

=2(a_n+1-a_n+3)=2b_nb_n=a_n+1-a_n+3為公比為2的等比數(shù)列.

(2)解:a₂=2a₁-1=-3,b₁=a₂-a₁+3=1b_n=a_n+1-a_n+3=2^n-12a_n+3n-4-a_n+3=2^n-1

a_n=2^n-1-3n+1(n∈N^*).

(3)解:設(shè)數(shù)列{a_n}的前n項(xiàng)和為S_n,S_n=2ⁿ-1=2ⁿ-1,

T_n=|a₁|+|a₂|+…+|a_n|,∵n≤4,a_n＜0,n＞4,a_n＞0,

∴n≤4,T_n=-S_n=1+-2ⁿ,n＞4,T_n=S_n-2S₄=2ⁿ+21.