问题
解答题
数列{an}的首项a1=1,前n项和为Sn,满足关系3tSn-(2t+3)Sn-1=3t(t>0,n=2,3,4…) (1)求证:数列{an}为等比数列; (2)设数列{an}的公比为f(t),作数列{bn},使b1=1,bn=f(
(3)求Tn=(b1b2-b2b3)+(b3b4-b4b5)+…+(b2n-1b2n-b2nb2n+1)的值. |
答案
(1)证:∵3tSn-(2t+3)Sn-1=3t,3tSn+1-(2t+3)Sn=3t(n≥2),两式相减得3tan+1-(2t+3)an=0
又t>0
∴
=an+1 an
(n≥2),2t+3 3t
又当n=2时,3tS2-(2t+3)S1=3t,
即3t(a1+a2)-(2t+3)a1=3t,得a2=
,2t+3 3t
即
=a2 a1
,2t+3 3t
∴
=an+1 an
(n≥1),2t+3 3t
∴{an}为等比数列
(2)由已知得,f(t)=2t+3 3t
∴bn=f(
)=1 bn-1
=3+ 2 bn-1 3 bn-1
+bn-1(n≥2,n∈N*).2 3
∴{bn}是一个首项为1,公差为
的等差数列.2 3
于是bn=
n+2 3 1 3
(3)Tn=b1b2-b2b3+b3b4-…+b2n-1b2n-b2nb2n+1
=b2(b1-b3)+b4(b3-b5)+…+b2n(b2n-1-b2n+1)=-2(b2+b4+…+b2n)
=-2d(b2+b4+…+b2n)
=-2×
[2 3
+5n 3
×n(n-1) 2
]4 3
=-
-8n2 9 4n 3