Question

高一数列大题

已知数列{an}前几项和Sn=n^2:设bn=(1/3)^n*an,数列{bn}的前n项和为Tn
(1)求数列{an}的通项公式
（2）证明：Tn<1

Answer 1

(1)解：
当n=1时，a1=S1=1^2=1；
当n>1时，Sn=n^2，S(n-1)=(n-1)^2=n^2-2n+1
则an=Sn-S(n-1)=n^2-(n^2-2n+1)=2n-1.
因为a1=1=2*1-1，符合上式，所以求数列{an}的通项公式是an=2n-1.
(2)证明：bn=(1/3)^n*an=(2n-1)*(1/3)^n
则 Tn=1/3+3*(1/3)^2+5*(1/3)^3+…+(2n-1)*(1/3)^n
(1/3)Tn= (1/3)^2+3*(1/3)^3+…+(2n-3)*(1/3)^n+(2n-1)*(1/3)^(n+1)
两式相减得：
(2/3)Tn=1/3+2*(1/3)^2+2*(1/3)^3+…+2*(1/3)^n-(2n-1)*(1/3)^(n+1)
=1/3+2*(1/9)*[1-(1/3)^(n-1)]/(1-1/3)-(2n-1)*(1/3)^(n+1)
=1/3+(1/3)*[1-(1/3)^(n-1)]-(2n-1)*(1/3)^(n+1)
=2/3-(1/3)^n-(2n-1)*(1/3)^(n+1)
所以Tn=1-(3/2)*[(1/3)^n+(2n-1)*(1/3)^(n+1)]<1.

Answer 2

1.首先n=1时 a1=s1=1^2=1
n>1时，an=sn-s(n-1)=n^2-(n-1)^2=n^2-n^2+2n-1=2n-1
将1代入，发现成立
{an}的通项公式为an=2n-1
2.bn=1/3)^n*(2n-1)
Tn=1/3*1+(1/3)^2*3+(1/3)^3*5+.....+(1/3)^n*(2n-1)
1/3Tn= (1/3)^2*1+(1/3)^3*3+.....+(1/3)^n*(2n-3)+(1/3)^(n+1)*(2n-1)
相减，2/3Tn=1/3*1+2*[(1/3)^2+(1/3)^3+.....(1/3)^n]-(1/3)^(n+1)*(2n-1)
(1/3)^2+(1/3)^3+.....(1/3)^n=[(1/3)^2-(1/3)^n*(1/3)]/(1-1/3)
=[(1/3)^2-(1/3)^(n+1)]/(2/3)
=[(1/3)^2-(1/3)^(n+1)]*(3/2)
2*[(1/3)^2+(1/3)^3+.....(1/3)^n]=2*[(1/3)^2-(1/3)^(n+1)]*(3/2)
=(1/3)^1-(1/3)^n<1/3
2/3Tn=1/3*1+2*[(1/3)^2+(1/3)^3+.....(1/3)^n]-(1/3)^(n+1)*(2n-1)
又因为-(1/3)^(n+1)*(2n-1)<0
2/3Tn<2/3
Tn<1

Answer 3

⑴∵Sn=n^2∴ Sn-1=(n-1)^2(n≥2)∴an=Sn-Sn-1=2n-1(n≥2)a1=1满足∴an=2n-1
⑵∴bn=(1/3)^n*an=n×(1/3）^(2n-1)
Tn=T1+T2+T3+……+Tn
=1·(1/3)^1+2·(1/3)^3+……+n×(1/3）^(2n-1)①
∴(1/9)Tn= =1·(1/3)^3+2·(1/3)^5+……+n×(1/3）^(2n+1)②
①-②得（8/9）Tn=(1/3)^1+(1/3)^3+……+(1/3）^(2n-1)-n(1/3)^(2n+1)求出Tn后，简单判断一下，即可得到结果。

Answer 4

an=2+2n