设等差数列an公差为d,点(an,bn)在函数f(x)=2^x的图像上1 若a1=1,点(a8,4b1)在函数f(x)的图像上,求数列an的前n项和Sn2 若a1=1,函数f(x)的图像在点（a2,b2)处的切线在x轴上的截距为2-1/ln2,求数列an/bn的前n项

Answer 1

an=a1+(n-1)d
点(an,bn)在函数f(x)=2^x的图像上
bn=2^(an) = 2^(a1+(n-1)d ) (1)
(1)
a1=1
(a8,4b1)在函数f(x)的图像上
4b1= 2^(a8)
= 2^(1+7d) (2)
from (1)
put n=1
b1=2 (3)
sub (3) into (2)
8=2^(1+7d)
1+7d=3
d=-2/7
an = 1-(2/7)(n-1)
Sn = -(1/7)(n-16)n
(2)
a1=1
f(x) = 2^x
f'(x) = ln2 .2^x
f'(a2) = ln2 .2^(1+d)
equation of tangent at (a2,b2)
y-b2 = f'(a2) .(x-a2)
y- 2^(1+d) =ln2 .2^(1+d) .[ x- (1+d) ] (4)
点（a2,b2)处的切线在x轴上的截距为2-1/ln2
for (4),(2-1/ln2,0)
- 2^(1+d) =ln2 .2^(1+d) .[ 2-1/ln2- (1+d) ]
= 2^(1+d) .[ 2ln2-1 -(1+d)ln2 ]
2ln2-1 -(1+d)ln2 = -1
ln2 -dln2 =0
d=1
an= n
bn = 2^(an)= 2^n
let
S= 1.(1/2)^1+2.(1/2)^2+...+n.(1/2)^n (5)
(1/2)S= 1.(1/2)^2+2.(1/2)^3+...+n.(1/2)^(n+1) (6)
(5)-(6)
(1/2)S= (1/2+1/2^2+...+1/2^n)-n.(1/2)^(n+1)
= (1- (1/2)^n) -n.(1/2)^(n+1)
S = 2(1- (1/2)^n) -n.(1/2)^n
= 2 -(n+2)(1/2)^n
cn = an/bn
= n.(1/2)^n
Tn = c1+c2+...+cn
= S
=2 -(n+2)(1/2)^n