Question

已知數(shù)列{a_n}的前三項(xiàng)與數(shù)列{b_n}的前三項(xiàng)對(duì)應(yīng)相同,且a₁+2a₂+2²a₃+…+2^n-1a_n=8n對(duì)任意的n∈N₊都成立,數(shù)列{b_n+1-b_n}是等差數(shù)列.

(1)求數(shù)列{a_n}與{b_n}的通項(xiàng)公式;

(2)問是否存在k∈N₊,使得b_k-a_k∈(0,1)?請(qǐng)說明理由.

Answer 1

解析：(1)已知a₁+2a₂+2²a₃+…+2^n-1a_n=8n(n∈N₊)             ①,

當(dāng)n≥2時(shí)，a₁+2a₂+2²a₃+…+2^n-2a_n-1=8(n-1)(n∈N₊)            ②,

①-②得，2^n-1a_n=8，求得a_n=2^4-n,在①中令n=1，可得a₁=8=2^4-1,

∴a_n=2^4-n(n∈N₊).

由題意知b₁=8,b₂=4,b₃=2,

∴b₂-b₁=-4,b₃-b₂=-2,

∴數(shù)列{b_n+1-b_n}的公差為-2-(-4)=2,

∴b_n+1-b_n=-4+(n-1)×2=2n-6,

b_n=b₁+(b₂-b₁)+(b₃-b₂)+…+(b_n-b_n-1)

=8+(-4)+(-2)+…+(2n-8)

=n²-7n+14(n∈N₊).

(2)b_k-a_k=k²-7k+14-2^4-k,

當(dāng)k≥4時(shí)f(k)=(k-)²+-2^4-k單調(diào)遞增，且f(4)=1,

∴

又f(1)=f(2)=f(3)=0,

∴不存在k∈N₊，使得b_k-a_k∈(0,1).