已知函数f(x)=ax-1(a不等于0),且f(2)f(5)f(14)成等比数列,设an=f(n)(

题目

已知函数f(x)=ax-1(a不等于0),且f(2)f(5)f(14)成等比数列,设an=f(n)(
n属于N*)
一：求数列{an}(n在右下角的)的前n项和Sn(n在右角)下.
二：设bn=2n(n在左上角的),求数列{anbn}的前n项和Tn

答案

f(x)=ax-1(a不等于0),且f(2)f(5)f(14)成等比数列,设an=f(n)(n属于N*),
令,a1=f(2),a2=f(5),a3=f(14).有
a1=2a-1,a2=5a-1,a3=14a-1.
f(2)f(5)f(14)成等比数列,
(a2)^2=a1*a3,
(5a-1)^2=(2a-1)(14a-1),
a(a-2)=0,(a≠),
a=2.
a1=f(2)=3,a2=f(5)=9,a3=f(14)=27.
q=a2/a1=3,
∴Sn=a1(1-q^n)/(1-q)=3(1-3^n)(1-3)=-3/2(1-3^n).
bn=2^n,数列{anbn}有,
{a1b1}={3*2}=6,
{a2b2}={9*2^2}=36,
{a3b3}={27*2^3}=216.
{a2b2}/{a1b1}=36/6=6=q2
数列{anbn}是公比为q2=6,的等比数列,有
Tn={a1b1}[1-(q2)^n]/(1-q2)
=6[1-6^n]/(1-6)
=-6/5(1-6^n).