问题 解答题
设函数f(x)=
2x+3
3x
,作数列{bn}:b1=1,bn=f(
1
bn-1
)(n≥2)

求和:Wn=b1b2-b2b3+b3b4-…+(-1)n-1bnbn+1
答案

∵f(x)=

2x+3
3x
=
2
3
+
1
x
bn=f(
1
bn-1
),n≥2

bn=

2
3
+bn-1

∵b1=1,∴{bn}是首项为1,公差为

2
3
的等差数列,

bn=

2n+1
3

bnbn+1=

1
9
(4n2+8n+3),

①当n为偶数时:

b2=

5
3
bn=
2n+1
3

∴Wn=b1b2-b2b3+b3b4-b4b5+…+bn-1bn-bnbn+1

=b2(b1-b3)+b4(b3-b5)+…+bn(bn-1-bn+1

=-2×

2
3
(b2+b4+…+bn

=-

4
3
×[
n
4
(
5
3
+
2n+1
3
)]

=-

1
9
(2n2+6n);

②当n为奇数时:

b2=

5
3
bn-1=
2n-1
3

∴Wn=b1b2-b2b3+b3b4-b4b5+…+bn-2bn-1-bn-1bn+bnbn+1

=b2(b1-b3)+b4(b3-b5)+…+bn-1(bn-2-bn)+bnbn+1

=-2×

2
3
(b2+b4+…+bn-1)+bnbn+1

=-

4
3
×[
n-1
4
(
5
3
+
2n-1
3
)
]+
1
9
(4n2+8n+3)

=

1
9
(2n2+6n+7).

Wn=

-
1
9
(2n2+6n),n为偶数
1
9
(2n2+6n+7),n为奇数

单项选择题
单项选择题