Question

数列｛an｝的前n项和记为Sn,a1＝t,点(Sn,an+1)在直线y=2x+1上，n∈N

(1)当实数t为何值时{an}是等比数？

(2)在(1)的结论下，设bn=log3an+1,Cn=bnbn+2分之1求数列｛Cn｝的前n项的和Tn

Answer 1

题干应该是: 数列{a_n}的首项a_1=t, 前n项和记为S_n. 已知点(S_n, a_{n+1})在直线y=2x+1上.

(1) 由已知, a_{n+1}=2S_n+1, 则a_n=2S_{n-1}+1, n>1. 两式相减得到a_{n+1}-a_n=2a_n, 即得a_{n+1}=3a_n, n>1. 但a_2=2S_1+1=2t+1, 所以要使得{a_n}成为等比数列, 必有a_2=3a_1, 即2t+1=3t, 解得t=1.
经验证, t=1满足题意且a_n=3^{n-1} (n>=1), {a_n}是公比为3的等比数列.
(2) 由(1)的解答可知a_n=3^{n-1}, 则b_n=n (n>=1), c_n=1/[n(n+2)]=1/2[1/n-1/(n+2)]. 于是
T_n=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+1)-1/(n+2)]
=n(3n+5)/[4(n+1)(n+2)].

Answer 2

点(Sn,an+1)在直线y=2x+1上
an+1=2Sn +1
an=2Sn-1 +1
将上两式相减得
an+1 - an = 2(Sn - Sn-1) = 2an
an+1=3 an
an=a1×3^(n-1) = t×3^(n-1)
S2=a1+a2=t+3t=4t
a3=a1*3*3=9t
a3=2S2 +1
9t=8t+1
t=1
∴　当实数t=1时,{an}是等比数列

bn=log(3) an+1= log(3) 3^n=n
Cn=bn*bn+2=n*(n+2)=n²+2n
Tn=C1+C2+……Cn
=(1+4+9+……+n²) + 2(1+2+3+……+n)
=n(n+1)(2n+1)/6 + 2n(n+1)/2
=n(n+1)(2n+7)/6