等差数列｛an｝的各项均为正数,a1=3,前n项和为Sn,｛bn｝为等比数列,b1=2,且b2S2=32,b3S3=120（1）求an与bn.（2）求数列｛anbn｝的前n项和Tn.（3）若(1/S1)+(1/S2)+…(1/Sn)≤x²+ax+1对任意正整数n和任意x

等差数列｛an｝的各项均为正数,a1=3,前n项和为Sn,｛bn｝为等比数列,b1=2,且b2S2=32,b3S3=120（1）求an与bn.（2）求数列｛anbn｝的前n项和Tn.（3）若(1/S1)+(1/S2)+…(1/Sn)≤x²+ax+1对任意正整数n和任意x
等差数列｛an｝的各项均为正数,a1=3,前n项和为Sn,｛bn｝为等比数列,b1=2,且b2S2=32,b3S3=120
（1）求an与bn.
（2）求数列｛anbn｝的前n项和Tn.
（3）若(1/S1)+(1/S2)+…(1/Sn)≤x²+ax+1对任意正整数n和任意x∈R恒成立,求实数a的取值范围.

等差数列｛an｝的各项均为正数,a1=3,前n项和为Sn,｛bn｝为等比数列,b1=2,且b2S2=32,b3S3=120（1）求an与bn.（2）求数列｛anbn｝的前n项和Tn.（3）若(1/S1)+(1/S2)+…(1/Sn)≤x²+ax+1对任意正整数n和任意x
(1) b2*S2=(b1*q)*(a1+a1+d)=2q*(6+d)=32 => q(6+d)=16
b3*S3=(b1*q^2)*(3a1+3d)=2q^2*(9+3d)=120 => q^2*(3+d)=20
联立解得：d=2 ,q=2（因为等差数列｛an｝的各项均为正数则d>0,故d= -6/5 ,q=10/3舍掉）
则：(1) an=3+2(n-1)=2n+1
bn=2*2^(n-1)=2^n
(2) an*bn=(2n+1)*2^n
则Tn=3*2+5*2^2+.+(2n+1)*2^n
(1/2)Tn=3+5*2+.+(2n+1)*2^(n-1)
(1/2)Tn-Tn=3+2*2+2*2^2+.+2*2^(n-1)-(2n+1)*2^n
(-1/2)Tn=3+2*2*[1-2^(n-1)]/(1-2)-(2n+1)*2^n
=3+2^(n+1)-4-(2n+1)*2^n
Tn=2-2*2^(n+1)+(2n+1)*2^(n+1)
=2+(2n-1)*2^(n+1)
(3)Sn为等差数列｛an｝的前n项和,则Sn=n*a1+n(n-1)/2=n*(n+2)
故1/Sn=1/[n*(n+2)]=1/2[1/n-1/(n+2)]
则：(1/S1)+(1/S2)+…(1/Sn)=1/2[(1-1/3)+(1/2-1/4)+(1/3-1/5)+…+(1/n)-1/(n+2)]
=1/2[1+1/2-1/(n+2)]
=3/4-1/(2n+4)

设a2=a1+m=3+m
则a3=a2+m=3+2m
S2=6+m,S3=9+3m
设b2=b1*k=2k, 则b3=2k^2
b2S2=2k(6+m)=32
b3S3=2k^2(9+3m)=120
求得m=2, k=2
an=3+2(n-1)
bn=2^n
剩下的你会了。

b2*S2=(b1*q)*(a1+a1+d)=2q*(6+d)=32
所以q(6+d)=16 (1)
b3*S3=(b1*q^2)*(3a1+3d)=2q^2*(9+3d)=120
所以q^2*(3+d)=20 (2)
联立(1)(2)解得 d=2 q=2
(1) an=3+2(n-1)=2n+1<...

全部展开

b2*S2=(b1*q)*(a1+a1+d)=2q*(6+d)=32
所以q(6+d)=16 (1)
b3*S3=(b1*q^2)*(3a1+3d)=2q^2*(9+3d)=120
所以q^2*(3+d)=20 (2)
联立(1)(2)解得 d=2 q=2
(1) an=3+2(n-1)=2n+1
bn=2*2^(n-1)=2^n
(2) an*bn=(2n+1)*2^n
所以Tn=3*2+5*2^2+.....+(2n+1)*2^n
(1/2)Tn=3+5*2+....+(2n+1)*2^(n-1)
(1/2)Tn-Tn=3+2*2+2*2^2+.....+2*2^(n-1)-(2n+1)*2^n
(-1/2)Tn=3+2*2*[2^(n-1)-1]/(2-1)-(2n+1)*2^n
=3+2^(n+1)-4-(2n+1)*2^n
Tn=2-2*2^(n+1)+(2n+1)*2^(n+1)
=2+(2n-1)*2^(n+1)
(3) Sn=(3+2n+1)*n/2=n(n+2)
1/Sn=1/n(n+2)=(1/2)[1/n-1/(n+2)]
所以(1/S1)+(1/S2)+…(1/Sn)
=(1/2)*[(1-1/3)+(1/2-1/4)+....+1/n-1/(n+2)]
=(1/2)*[3/2-1/(n+1)-1/(n+2)]
≤x²+ax+1恒成立
只需x²+ax+1≥3/4
x²+ax+1/4≥0
则需判别式=a²-4*(1/4)≤0
解得-1≤a≤1

收起

ddd