Question

已知点pn(an,bn)满足a n+1=anb n+1,b n+1=bn/1-4an^2,且p1的坐标为（1,-1）.

Answer 1

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

又bn=an+1-an=an+k,则bn+1=an+1+k,

所以==2,故数列{bn}是等比数列.

(2)由(1),b1=a1+k,bn=b1×2n-1=(a1+k)2n-1,an=bn-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.