Question

设{an}为等差数列，{bn}为等比数列，已知a1=b1=1,a2+a4=b3,b2b4=a3,分别求出{an}及{bn}前10项的和S10及T10

Answer 1

a2+a4=b3
2a3=b3
a3=b3/2
b2b4=a3
(b3)^2=a3
(b3)^2=b3/2
b3=1/2
a3=(1/2)/2=1/4

a3=a1+2d
1/4=1+2d
d=-3/8
Sn=[2a1+(n-1)d]*n/2
=[2*1+9*(-3/8)]*10/2
=[2-27/8]*5
=-55/8

b3=b1q^2
1/2=q^2
q=±√2/2

当q=√2/2时
T10=b1(1-q^n)/(1-q)
=[1-(√2/2)^10]/(1-√2/2)
=[1-1/32]/[(2-√2)/2]
=(31/16)/(2-√2)
=(31/16)(2+√2)/(2-√2)(2+√2)
=(31/16)(2+√2)/2
=31(2+√2)/32
=(62+31√2)/32

当q=-√2/2时
T10=b1(1-q^n)/(1-q)
=[1-(-√2/2)^10]/(1+√2/2)
=[1-1/32]/[(2+√2)/2]
=(31/16)/(2+√2)
=(31/16)(2-√2)/(2-√2)(2+√2)
=(31/16)(2-√2)/2
=31(2-√2)/32
=(62-31√2)/32

Answer 2

a1=b1=1
a2=1+d
a4=1+3d
a3=1+2d
b3=q^2
b2=q
b4=q^3
所以1+d+1+3d=q^2,2+4d=q^2
q^4=1+2d
相除
(2+4d)/(1+2d)=q^2/q^4
q^2=1/2
d=(q^2-2)/4=-3/8
q=±√2/2

S10=(a1+a10)*10/2=(a1+a1+9d)*10/2=(2-27/8)*5=-55/8
T10=b1*(1-q^10)/(1-q)=1*[1-(1/2)^5]/(1±√2/2)=(62±31√2)/32

Answer 3

a2+a4=b3
2a3=b3
a3=b3/2
b2b4=a3
(b3)^2=a3
(b3)^2=b3/2
b3=1/2
a3=(1/2)/2=1/4

a3=a1+2d
1/4=1+2d
d=-3/8
Sn=[2a1+(n-1)d]*n/2
=[2*1+9*(-3/8)]*10/2
=[2-27/8]*5
=-55/8

b3=b1q^2
1/2=q^2
q=±√2/2

当q=√2/2时
T10=b1(1-q^n)/(1-q)
=[1-(√2/2)^10]/(1-√2/2)
=[1-1/32]/[(2-√2)/2]
=(31/16)/(2-√2)
=(31/16)(2+√2)/(2-√2)(2+√2)
=(31/16)(2+√2)/2
=31(2+√2)/32
=(62+31√2)/32

当q=-√2/2时
T10=b1(1-q^n)/(1-q)
=[1-(-√2/2)^10]/(1+√2/2)
=[1-1/32]/[(2+√2)/2]
=(31/16)/(2+√2)
=(31/16)(2-√2)/(2-√2)(2+√2)
=(31/16)(2-√2)/2
=31(2-√2)/32
=(62-31√2)/32