数列{an}中,an=1+2+3+…+nn,bn=1anan+1的前n项和为______.-数学

题目简介

数列{an}中,an=1+2+3+…+nn,bn=1anan+1的前n项和为______.-数学

题目详情

数列{an}中,an=
1+2+3+…+n
n
bn=
1
anan+1
的前n项和为______.
题型:填空题难度:中档来源:不详

答案

设数列bn的前n项和为Sn
由题意可得an=class="stub"1+2+3+…+n
n
=
n(n+1)
2
n
=class="stub"n+1
2

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

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

∴Sn=b1+b2+…+bn-1+bn
=4(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
+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

bn=class="stub"1
anan+1
的前n项和为class="stub"2n
n+2

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