Question

已知函数f(x)=x^2-4x+2,数列{an}是等差数列，且a1=f(x+1),a2=0, a3=f(x-1)，若bn=（根号2）的an次方，

求数列｛an*bn｝的前n项和

Answer 1

f(x)=x^2-4x+2
a1=f(x+1)
= x^2-2x-1
a2=0
a3=f(x-1)
=x^2-6x+7
a1+a3= 2a2
x^2-2x-1+x^2-6x+7=0
x^2-4x+3 =0
(x-1)(x-3)=0
x=1 or 3

case 1: x=1
a1=f(2)= -2
a3=f(0) = 2
an = -2+2(n-1) = 2n-4
bn=(√2)^an
=(√2)^(2n-4)
= 2^(n-2)
an.bn = (2n-4).2^(n-2)
= n. 2^(n-1) - 2^n
Sn = a1b1+a2b2+...+anbn
= [∑(i:1->n) i.2^(i-1)] - 2(2^n-1)
consider
1+x+x^2+...+x^n = (x^(n+1) -1)/(x-1)
1+2x+...+nx^(n-1) = [(x^(n+1) -1)/(x-1)]'
= (nx^(n+1) -(n+1)x^n +1)/(x-1)^2
put x=2
∑(i:1->n) i.2^(i-1) = [n.2^(n+1) -(n+1).2^n +1]
= 1+ (n-1).2^n
Sn = a1b1+a2b2+...+anbn
= [∑(i:1->n) i.2^(i-1)] - 2(2^n-1)
=1+ (n-1).2^n - 2(2^n-1)
= 3 +(n-3).2^n

case 2: x=3
a1=f(4) = 2
a3=f(2)= -2
an = 2+(n-1)(-2) = -2n+4

bn=(√2)^an
=(√2)^(-2n+4)
= 2^(-n+2)
an.bn = (-2n+4).2^(-n+2)
= -n.(2)^(-n+3) + 2^(-n+4)
= -4[n. (1/2)^(n-1)] +2^(-n+4)
Sn = a1b1+a2b2+...+anbn
= -4[∑(i:1->n) i.(1/2)^(i-1)] + 16( 1- (1/2)^n)
consider
1+x+x^2+...+x^n = (x^(n+1) -1)/(x-1)
1+2x+...+nx^(n-1) = [(x^(n+1) -1)/(x-1)]'
= (nx^(n+1) -(n+1)x^n +1)/(x-1)^2
put x=1/2
∑(i:1->n) i.(1/2)^(i-1) = 4[n.(1/2)^(n+1) -(n+1).(1/2)^n +1]
= 4[1- (n+2).(1/2)^(n+1)]
Sn = a1b1+a2b2+...+anbn
= -4[∑(i:1->n) i.(1/2)^(i-1)] + 16( 1- (1/2)^n)
= -16[1- (n+2).(1/2)^(n+1)] +16( 1- (1/2)^n)
=(16n)(1/2)^(n+1)
= 8n. (1/2)^n