(文)数列{an}满足an+1=n+2nan(n∈N*),且a1=1.(1)求通项an;(2)记bn=1an,数列{bn}的前n项和为Sn,求Sn.-数学

题目简介

(文)数列{an}满足an+1=n+2nan(n∈N*),且a1=1.(1)求通项an;(2)记bn=1an,数列{bn}的前n项和为Sn,求Sn.-数学

题目详情

(文)数列{an}满足an+1=
n+2
n
an
(n∈N*),且a1=1.(1)求通项an;(2)记bn=
1
an
,数列{bn}的前n项和为Sn,求Sn
题型:解答题难度:中档来源:甘谷县模拟

答案

解(1)∵an+1=class="stub"n+2
n
an

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

∵a1=1
a2
a1
=class="stub"3
1
a3
a2
=class="stub"4
2
an
an-1
=class="stub"n+1
n-1

以上n-1个式子相乘可得,
a2
a1
a3
a2
an
an-1
=class="stub"3
1
×class="stub"4
2
×class="stub"5
3
…class="stub"n-1
n-3
×class="stub"n
n-2
×class="stub"n+1
n-1

an
a1
=
n(n+1)
1×2

∴an=
n(n+1)
2

(2)∵bn=class="stub"1
an
=class="stub"2
n(n+1)
=2(class="stub"1
n
-class="stub"1
n+1
)

Sn=2(1-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
+…+class="stub"1
n
-class="stub"1
n+1
)=2(1-class="stub"1
n+1
)=class="stub"2n
n+1

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