已知各项均为正数的数列﹛an﹜,对于任意正整数n,点(an,sn)在曲线y=12(x2+x)上(1)求证:数列﹛an﹜是等差数列;(2)若数列﹛bn﹜满足bn=1an•an+2,求数列﹛bn﹜的前n项

题目简介

已知各项均为正数的数列﹛an﹜,对于任意正整数n,点(an,sn)在曲线y=12(x2+x)上(1)求证:数列﹛an﹜是等差数列;(2)若数列﹛bn﹜满足bn=1an•an+2,求数列﹛bn﹜的前n项

题目详情

已知各项均为正数的数列﹛an﹜,对于任意正整数n,点(an,sn)在曲线y=
1
2
(x2+x)

(1)求证:数列﹛an﹜是等差数列;
(2)若数列﹛bn﹜满足bn=
1
anan+2
,求数列﹛bn﹜的前n项和Tn
题型:解答题难度:中档来源:不详

答案

(1)∵各项均为正数的数列﹛an﹜,对于任意正整数n,点(an,sn)在曲线y=class="stub"1
2
(x2+x)
上,
Sn=class="stub"1
2
(an2+an)
,①
∴Sn-1=class="stub"1
2
an-12+an-1),n≥2,②
①-②,得an=Sn-Sn-1=class="stub"1
2
[(an2+an)-(an-12+an-1)]
an-12+an-1=an2-an
an2-an-12=an+an-1,
∴an-an-1=1.
∴数列﹛an﹜是等差数列.
(2)∵Sn=class="stub"1
2
(an2+an)

a1=class="stub"1
2
(a12+a1)
,解得a1=1,a1=0(舍),
∵an-an-1=1.
∴an=1+(n-1)=n,
∴bn=class="stub"1
anan+2
=class="stub"1
n(n+2)
=class="stub"1
2
(class="stub"1
n
-class="stub"1
n+2
)

∴数列﹛bn﹜的前n项和
Tn=b1+b2+b3+…+bn
=class="stub"1
2
(1-class="stub"1
3
)+class="stub"1
2
(class="stub"1
2
-class="stub"1
4
)
+class="stub"1
2
class="stub"1
3
-class="stub"1
5
)+…+class="stub"1
2
(class="stub"1
n
-class="stub"1
n+2
)

=class="stub"1
2
(1+class="stub"1
2
-class="stub"1
n+1
-class="stub"1
n+2

=class="stub"3
4
-class="stub"1
2n+2
-class="stub"1
2n+4

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