题目内容

设{an}是正数组成的数列,前n项和为Snan=2
2Sn
-2

(Ⅰ)写出数列{an}的前三项;
(Ⅱ)求数列{an}的通项公式,并写出推证过程;
(Ⅲ)令bn=
4
anan+1
,求数列{bn}的前n项和Tn
(Ⅰ)∵an=2
2Sn
-2

n=1时可得,a1=2
2s1
-2
∴a1=2
把n=2代入可得a2=6,n=3代入可得a3=10;
(Ⅱ)8Sn=an2+4an+4…(1)
8Sn+1=an+12+4an+1+4…(2)
(2)-(1)得8an+1=an+12-an2+4an+1-4an
(an+1+an)(an+1-an-4)=0
∵an+1+an>0
∴an+1-an-4=0
an+1-an=4
∴{an}是以2为首项,4为公差的等差数列.an=a1+(n-1)d=4n-2
( III)bn=
4
anan+1
=
4
(4n-2)(4n+2)
=
1
(2n-1)(2n+1)
=
1
2
(
1
2n-1
-
1
2n+1
)

∴Tn=b1+b2+…+bn
=
1
2
(1-
1
3
+
1
3
-
1
5
+…+
1
2n-1
-
1
2n+1
)

=
1
2
(1-
1
2n+1
)=
n
2n+1
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