Question

1.数列{An}的前n项和是Sn,满足Sn=(3/2)An-n/2-3/4,设Bn=log3 (An+1/2),则数列{

Answer 1

1.Sn=(3/2)An-n/2-3/4

Sn-1=(3/2)An-1 - (n-1)/2-3/4

两式相减,整理an=3an-1+1

an+1/2=3(an-1 + 1/2)

Sn=(3/2)An-n/2-3/4 n=1时 s1=a1=(3/2)a1-1/2-3/4

a1=5/2 a1+1/2=3

an+1/2=3*3^(n-1)=3^n

Bn=Log（3^n）=n

所以数列是 1/n(n+1)=1/n-1/n+1

T19=1/1-1/2+1/2.1/19-1/20=19/20

2.显然圆心到F和到直线距离相等 由定义,轨迹是抛物线X^2=8y

设lab y-2=kx

联立两式 x^2-8kx-16=0

x1*x2=-16

可以用导数简单点,..Y=x^2/8 dy/dx=x/4

kAQ*kBQ=x1/4 * x2/4=x1*x2/16=-1

所以垂直..