数列{an}中,a1=1,a2=23,且1an-1+1an+1=2an.(1)求an;(2)设bn=anan+1,求b1+b2+b3+…bn;(3)求证:a12+a22+a32+…+an2<4-数学

题目简介

数列{an}中,a1=1,a2=23,且1an-1+1an+1=2an.(1)求an;(2)设bn=anan+1,求b1+b2+b3+…bn;(3)求证:a12+a22+a32+…+an2<4-数学

题目详情

数列{an}中,a1=1,a2=
2
3
,且
1
an-1
+
1
an+1
=
2
an

(1)求an
(2)设bn=anan+1,求b1+b2+b3+…bn
(3)求证:a12+a22+a32+…+an2<4
题型:解答题难度:中档来源:不详

答案

(1)依题意知{class="stub"1
an
}
为等差数列,公差d=class="stub"1
a2
-class="stub"1
a1
=class="stub"1
2

class="stub"1
an
=1+class="stub"1
2
(n-1)
,∴an=class="stub"2
n+1

(2)bn=anan+1=class="stub"4
(n+1)(n+2)
4(class="stub"1
n+1
-class="stub"1
n+2
)

∴b1+b2+…+bn=4[(class="stub"1
2
-class="stub"1
3
)+(class="stub"1
3
-class="stub"1
4
)+…+(class="stub"1
n+1
-class="stub"1
n+2
)]
=4(class="stub"1
2
-class="stub"1
n+2
) =class="stub"2n
n+2

(3)an2=class="stub"4
(n+1)2
<class="stub"4
n(n+1)
=4(class="stub"1
n+1
-class="stub"1
n+2
),
∴a12+a22+…+an2<4[(1-class="stub"1
2
) +(class="stub"1
2
-class="stub"1
3
)+…+(class="stub"1
n+1
-class="stub"1
n+2
)]
=4(1-class="stub"1
n+1
)<4.

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