题目内容
已知各项为正数的数列{an}的前n项和为Sn,满足Sn+1=(Ⅰ)求an;
(Ⅱ)求bn;
(Ⅲ)令cn=
,数列{cn}的前n项和为Tn,求证:Tn<3.
解:(Ⅰ)令n=1,a1+1=
,
∵a1+1>0,∴1=
,a1=2
由已知得2Sn+2=
+an①
2Sn+1+2=
+an+1②
②-①得(an+1+an=
-
,∵an>0,∴an+1-an=1
∴数列{an}是以2为首项,以1为公差的等差数列,
∴an=n+1
(Ⅱ)由bn+2+2bn=3bn+1得bn+2-bn+1=2(bn+1-bn)
∵b1=2,b2=4,b2- b1=2,∴
=2
∴数列{bn+1-bn}是以2为首项,以2为公比的等比数列
bn+1-bn=2n
b2-b1=2
b3-b2=22
b4-b3=23
……
bn-bn-1=2n-1
累加得bn=2n
(Ⅲ)cn=
=(n+1)![]()
Tn=2×
+3(
)2+4(
)3+…+(n+1)(
)n③
Tn=2(
)2+3(
)3+…+n(
)n+(n+1)(
)n+1④
③-④并整理得Tn=3-
<3
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