在数列{an}中,an=1n+1+2n+1+…+nn+1,又bn=2an•an+1,求数列{bn}的前n项的和.-数学

题目简介

在数列{an}中,an=1n+1+2n+1+…+nn+1,又bn=2an•an+1,求数列{bn}的前n项的和.-数学

题目详情

在数列{an}中,an=
1
n+1
+
2
n+1
+…+
n
n+1
,又bn=
2
anan+1
,求数列{bn}的前n项的和.
题型:解答题难度:中档来源:不详

答案

∵1+2+…+n=class="stub"1
2
n(n+1)

an=class="stub"1
n+1
+class="stub"2
n+1
+…+class="stub"n
n+1
=class="stub"n
2

bn=class="stub"2
class="stub"n
2
•class="stub"n+1
2
=8(class="stub"1
n
-class="stub"1
n+1
)

∴数列{bn}的前n项和Sn=8[(1-class="stub"1
2
)+(class="stub"1
2
-class="stub"1
3
)+(class="stub"1
3
-class="stub"1
4
)+…+(class="stub"1
n
-class="stub"1
n+1
)]
=8(1-class="stub"1
n+1
)
=class="stub"8n
n+1

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