已知函数,f(x)=x3x+1,数列{an}满足a1=1,an+1=f(an)(n∈N*)(I)求证数列{1an}是等差数列,并求数列{an}的通项公式;(II)记Sn=a1a2+a2a3+..ana

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*)(I)求证数列{1an}是等差数列,并求数列{an}的通项公式;(II)记Sn=a1a2+a2a3+..ana

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*
(I)求证数列{
1
an
}是等差数列,并求数列{an}的通项公式;
(II)记Sn=a1a2+a2a3+..anan+1,求Sn
题型:解答题难度:中档来源:许昌三模

答案

(I)由条件得,an+1=
an
3an+1

class="stub"1
an+1
=class="stub"1
an
+3
class="stub"1
an+1
-class="stub"1
an
=3.
∴数列{class="stub"1
an
}是首项为class="stub"1
a1
=1,公差d=3的等差数列.
class="stub"1
an
=1+(n-1)×3=3n-2.
故an=class="stub"1
3n-2

(II)∵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
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

更多内容推荐