数列{an}的前n项和为Sn,且满足a1=1,2Sn=(n+1)an,(I)求an与an-1的关系式,并求{an}的通项公式;(II)求和Wn=1a22-1+1a23-1+…+1a2n+1-1.-数学

题目简介

数列{an}的前n项和为Sn,且满足a1=1,2Sn=(n+1)an,(I)求an与an-1的关系式,并求{an}的通项公式;(II)求和Wn=1a22-1+1a23-1+…+1a2n+1-1.-数学

题目详情

数列{an}的前n项和为Sn,且满足a1=1,2Sn=(n+1)an
(I)求an与an-1的关系式,并求{an}的通项公式;
(II)求和Wn=
1
a22
-1
+
1
a23
-1
+…+
1
a2n+1
-1
题型:解答题难度:中档来源:不详

答案

(I)由已知
2Sn=(n+1)an
2Sn-1=nan-1
两式相减得2an=(n+1)an-nan-1,移向整理得出an=class="stub"n
n-1
an-1
(n≥2)

an
a1
=
an
an-1
an-1
an-2
•…•
a2
a1
=class="stub"n
n-1
•class="stub"n-1
n-2
•…•class="stub"2
1
=n

∴an=n;且a1=1也适合,
所以an=n.
(II)class="stub"1
a2n+1
-1
=class="stub"1
(n+1)2-1
=class="stub"1
n(n+2)
=class="stub"1
2
(class="stub"1
n
-class="stub"1
n+2
)

Wn=class="stub"1
1•3
+class="stub"1
2•4
+class="stub"1
3•5
+…+class="stub"1
n(n+2)
=class="stub"1
2
[(1-class="stub"1
3
)+(class="stub"1
2
-class="stub"1
4
)
+(class="stub"1
3
-class="stub"1
5
)+…+(class="stub"1
n-1
-class="stub"1
n+1
)+( 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"2n+3
2n(n+1)

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