Question

以数列{a_n}的任意相邻的两项为坐标的点P_n(a_n,a_n+1)(n∈N^*)均在一次函数y=2x+k的图象上,数列{b_n}满足条件:b_n=a_n+1-a_n(n∈N^*，b₁≠0).

(1)求证:数列{b_n}是等比数列;

(2)设数列{a_n}、{b_n}的前n项和分别为S_n、T_n,若S₆=T₄,S₅=-9,求k的值.

Answer 1

解析:(1)P_n(a_n,a_n+1)(n∈N^*)在一次函数y=2x+k的图象上,a_n+1=2a_n+k,即a_n+1+k=2(a_n+k),?

又b_n=a_n+1-a_n=a_n+k,则b_n+1=a_n+1+k,?

所以==2,故数列{b_n}是等比数列.?

(2)由(1),b₁=a₁+k,b_n=b₁×2^n-1=(a₁+k)2^n-1,a_n=b_n-k,?

则S₆=T₆-6k=-6k=63a₁+57k,?

T₄==15(a₁+k),?

由S₆=T₄得a₁=-k.?

又S₅=-9,即T₅-5k=-9,-5k=-9,?

即31a₁+26k=-9.?

将a₁=-k代入,得k=8.