Question

设数列{a_n}和{b_n}满足:a₁=b₁=6,a₂=b₂=4,a₃=b₃=3,且数列{a_n+1-a_n}(n∈N^*)是等差数列,数列{b_n-2}(n∈N^*)是等比数列.

(1)求列数{a_n}和{b_n}的通项公式.

(2)是否存在k∈N^*,使a_k-b_k∈(0,)?若存在,求出k;若不存在,请说明理由.

Answer 1

解:(1)由已知a₂-a₁=-2,a₃-a₂=-1,d=-1-(-2)=1,?

∴a_n+1-a_n=(a₂-a₁)+(n-1)×1=n-3.                                                                       ?

∴a_n=a₁+(a₂-a₁)+(a₃-a₂)+…+(a_n-a_n-1)?

=6+(-2)+(-1)+0+1+2+…+(n-4)?

=.                                                                                                      ?

由已知b₁-2=4,b₂-2=2,即q==,?

∴b_n-2=(b₁-2)·()^n-1=4·()^n-1=8·()ⁿ.                                                            ?

∴b_n=2+8·()ⁿ.                                                                                                      ?

(2)设f(k)=a_k-b_k=k²-k-8·()^k+7.?

当k≥4时, k²-k是k的增函数；-8·()^k也是k的增函数.?

∵f(4)= ,∴k≥4时,f(k)≥.                                                                           ?

∵f(1)=f(2)=f(3)=0,∴不存在k，使f(k)∈(0,).