已知函数f(x)=x3x+1,数列an满足a1=1,an+1=f(an)(n∈N*).(1)求数列{an}的通项公式;(2)记Sn=a1a2+a2a3+…+anan+1,求Sn.-数学

题目简介

已知函数f(x)=x3x+1,数列an满足a1=1,an+1=f(an)(n∈N*).(1)求数列{an}的通项公式;(2)记Sn=a1a2+a2a3+…+anan+1,求Sn.-数学

题目详情

已知函数f(x)=
x
3x+1
,数列an满足a1=1,an+1=f(an)(n∈N*).
(1)求数列{an}的通项公式;
(2)记Sn=a1a2+a2a3+…+anan+1,求Sn
题型:解答题难度:中档来源:不详

答案

(1)由已知得,an+1=
an
3an+1
,整理得class="stub"1
an+1
-class="stub"1
an
=3

∴数列{class="stub"1
an
}
是首项,公差的等差数列.
class="stub"1
an
=1+(n-1)×3=3n-2

an=class="stub"1
3n-2
(n∈N*)
(6分)
(2)∵anan+1=class="stub"1
(3n-2)(3n+1)
=class="stub"1
3
(class="stub"1
3n-2
-class="stub"1
3n+1
)

Sn=a1a2+a2a3+…+anan+1=class="stub"1
1×4
+class="stub"1
4×7
+…+class="stub"1
(3n-2)(3n+1)

=class="stub"1
3
[(1-class="stub"1
4
)+(class="stub"1
4
-class="stub"1
7
)+…+(class="stub"1
3n-2
-class="stub"1
3n+1
)]

=class="stub"1
3
(1-class="stub"1
3n+1
)=class="stub"n
3n+1
.(13分)

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