Question

高二数学求答案

Answer 1

8:a5=a3q²,a6=a4q²,a4=a3q
a3²(1+q²)+a3²q²(1+q²)=25
a3²(1+q²)(1+q²)=25
a3²(1+q²)²=25
a3(1+q²)=5
a3+a5=a3+a3q²=a3(1+q²)=5
9：等差数列
an=a1+(n-1)d=-60+3(n-1)=3n-63
an=0,3n-63=0,n=21
n<21,an<0,|an|递减，|an|=63-3n，n>21，|an|递增，|an|=3n-63。a30=90-63=27
|a1|+...+|a21|=(60+0)x21/2=630
|a21|+...+|a30|=(0+27)x10/2=135
630+135=765
10:
an=a1+(n-1)d1
bn=b1+(n-1)d2

Sn=n(a1+an)/2=n[2a1+(n-1)d1]/2
Tn=n(b1+bn)/2=n[2b1+(n-1)d2]/2
Sn/Tn=[2a1+(n-1)d1]/[2b1+(n-1)d2]=2n/(3n+1)

a1/b1=S1/T1=2/(3+1)=1/2
b1=2a1

S2/T2=[2a1+d1]/[2b1+d2]=[2a1+d1]/[4a1+d2]=2x2/(3x2+1)=4/7
7(2a1+d1)=4(4a1+d2)
14a1+7d1=16a1+4d2
d2=(7d1-2a1)/4

S3/T3=[2a1+2d1]/[2b1+2d2]=2(a1+d1)/[4a1+(7d1-2a1)/2]=2(a1+d1)/[3a1+7d1/2]=2x3/(3x3+1)=6/10
10(a1+d1)=3[3a1+7d1/2]
10a1+10d1=9a1+21d1/2
a1=d1/2
d1=2a1
d2=(7d1-2a1)/4=(14a1-2a1)/4=3a1
an=a1+(n-1)2a1=a1[1+2n-2]=a1(2n-1)
bn=b1+(n-1)d2=2a1+(n-1)3a1=a1[2+3n-3]=a1(3n-1)

an/bn=(2n-1)/(3n-1)

Sn=n[2a1+(n-1)d1]/2=n[2a1+(n-1)2a1]/2=na1[1+(n-1)]=n²a1
Tn=n[2b1+(n-1)d2]/2=n[4a1+(n-1)3a1]/2=na1[4+3n-3]/2=na1(3n+1)/2
Sn/Tn=n²/n(3n+1)/2=2n/(3n+1)

正确！

Answer 2

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