设{an}为等差数列,bn=（1/2)^an,b3b4b5=512,b1b5+2b3b5+b3b7=400,则{bn}的前n项和Sn=?

Question

设{an}为等差数列,bn=（1/2)^an ,b3b4b5=512,b1b5+2b3b5+b3b7=400,则{bn}______的前n项和Sn=______?

Answer 1

b3b4b5=512
(1/2)^a3*(1/2)^a4*(1/2)^a5*=512
(1/2)^(a3+a4+a5)=512
(1/2)^(3a4)=512
(1/2)^(3a4)=2^9
(1/2)^(3a4)=(1/2)^(-9)
3a4=-9
a4=-3
b4=(1/2)^a4
=(1/2)^-3
=2^3
=8
b1b5+2b3b5+b3b7=400
(b3)^2+2b3b5+(b5)^2=400
(b3+b5)^2=400
(b3+b5)^2=400
(b4/q+b4q)^2=400
(b4/q+b4q)^2=400
(8/q+8q)^2=400
64(1/q+q)^2=400
4(1/q+q)^2=25
4/q^2+8+4q^2=25
4/q^2-17+4q^2=0
4q^4-17q^2+4=0
(4q^2-1)(q^2-4)=0
(2q-1)(2q+1)(q-2)(q+2)=0
q=1/2或q=-1/2或q=2或q=-2
b3b4b5=512
b1q^2*b1q^3*b1^4=512
(b1)^3*q^9=512
当q=1/2时
(b1)^3*(1/2)^9=512
(b1)^3*(1/2)^9=(1/2)^-9
(b1)^3*=(1/2)^-18
b1=(1/2)^-6
sn=b1(1-q^n)/(1-q)
=(1/2)^(-6)*[1-(1/2)^n]/(1-1/2)
=(1/2)^(-6)*[1-(1/2)^n]/(1/2)
=(1/2)^(-7)*[1-(1/2)^n]
=(1/2)^(-7)-(1/2)^(n-7)
=128-(1/2)^(n-7)
当q=-1/2时
(b1)^3*(-1/2)^9=512
(b1)^3*(-1/2)^9=(1/2)^(-9)
(b1)^3*=-(1/2)^(-18)
b1=-(1/2)^(-6)
sn=b1(1-q^n)/(1-q)
=-(1/2)^(-6)*[1-(-1/2)^n]/(1+1/2)
=-(1/2)^(-6)*[1-(-1/2)^n]/(3/2)
=-(1/2)^(-7)*[1-(-1/2)^n]/3
=-(1/2)^(-7)+(-1/2)^(n-7)
=-128+(-1/2)^(n-7)
当q=2时
(b1)^3*2^9=512
(b1)^3*2^9=2^9
(b1)^3=1
b1=1
sn=b1(1-q^n)/(1-q)
=[1-2^n]/(1-2)
=2^n-1
当q=-2时
(b1)^3*(-2)^9=512
-(b1)^3*2^9=2^9
(b1)^3=-1
b1=-1
sn=b1(1-q^n)/(1-q)
=-1*[1-(-2)^n]/(1+2)
=[(-2)^n-1]/3