> 数学 >
已知数列{an}的通项公式an
1+2+…+n
n
bn
1
anan+1
,则数列{bn}的前n项和为______.
人气:485 ℃ 时间:2019-11-04 02:45:02
解答
an
1+2+…+n
n
=
n(n+1)
2n
=
n+1
2
,∴bn
1
anan+1
=
4
(n+1)(n+2)
=4(
1
n+1
1
n+2
)

∴数列{bn}的前n项和=4[(
1
2
1
3
)+(
1
3
1
4
)+…+(
1
n+1
1
n+2
)]

=4(
1
2
1
n+2
)
=
2n
n+2

故答案为
2n
n+2
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