> 数学 >
等差数列{an}中,a7=4,a19=2a9
(Ⅰ)求{an}的通项公式;
(Ⅱ)设bn=
1
nan
,求数列{bn}的前n项和Sn
人气:378 ℃ 时间:2019-08-18 01:41:59
解答
(I)设等差数列{an}的公差为d
∵a7=4,a19=2a9
a1+6d=4
a1+18d=2(a1+8d)

解得,a1=1,d=
1
2

an=1+
1
2
(n−1)
=
1+n
2

(II)∵bn
1
nan
=
2
n(n+1)
=
2
n
2
n+1

∴sn=2(1−
1
2
+
1
2
1
3
+…+
1
n
1
n+1
)

=2(1−
1
n+1
)
=
2n
n+1
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版