已知数列{an}的前n项和Sn满足:S1=10,当n≥2时,2Sn=(n+4)an.(1)求a2,a3的值;(2)求数列{an}的通项公式;(3)求1a2a3+1a3a4+…+1anan+1的值.-数

题目简介

已知数列{an}的前n项和Sn满足:S1=10,当n≥2时,2Sn=(n+4)an.(1)求a2,a3的值;(2)求数列{an}的通项公式;(3)求1a2a3+1a3a4+…+1anan+1的值.-数

题目详情

已知数列{an}的前n项和Sn满足:S1=10,当n≥2时,2Sn=(n+4)an
(1)求a2,a3的值;  
(2)求数列{an}的通项公式;
(3)求
1
a2a3
+
1
a3a4
+…+
1
anan+1
的值.
题型:解答题难度:中档来源:不详

答案

(1)∵a1=S1=10,由2Sn=(n+4)an
令n=2,得2S2=(2+4)a2,即a1+a2=6a2,
∴a2=5
令n=3,得2S3=(3+4)a3,即2(a1+a2+a3)=7a3,
∴a3=6
(2)∵2Sn=(n+4)an,2Sn-1=(n+3)an-1(n≥3)
两式相减,得2an=2(Sn-Sn-1)=(n+4)an-(n+3)an-1
an
an-1
=class="stub"n+3
n+2
(n≥3)

an=a1×
a2
a1
×
a3
a2
×…
an-1
an-2
an
an-1
=10•class="stub"5
10
•class="stub"6
5
•class="stub"7
6
…class="stub"n+3
n+2
=n+3
(n≥3)
n=2时也适合,n=1时,a1=10不适合
an=
10(n=1)
n+3(n≥2)

(3)当n≥2时,
class="stub"1
anan+1
=class="stub"1
(n+3)(n+4)
=class="stub"1
n+3
-class="stub"1
n+4

class="stub"1
a2a3
+class="stub"1
a3a4
+…class="stub"1
anan+1
=(class="stub"1
5
-class="stub"1
6
)+(class="stub"1
6
-class="stub"1
7
)+…+(class="stub"1
n+3
-class="stub"1
n+4
)
=class="stub"1
5
-class="stub"1
n+4

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