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

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