设函数f(x)=2x+33x,作数列{bn}:b1=1,bn=f(1bn-1)(n≥2),求和:Wn=b1b2-b2b3+b3b4-…+(-1)n-1•bnbn+1.-数学

题目简介

设函数f(x)=2x+33x,作数列{bn}:b1=1,bn=f(1bn-1)(n≥2),求和:Wn=b1b2-b2b3+b3b4-…+(-1)n-1•bnbn+1.-数学

题目详情

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
题型:解答题难度:中档来源:不详

答案

∵f(x)=class="stub"2x+3
3x
=class="stub"2
3
+class="stub"1
x
bn=f(class="stub"1
bn-1
),n≥2

bn=class="stub"2
3
+bn-1

∵b1=1,∴{bn}是首项为1,公差为class="stub"2
3
的等差数列,
bn=class="stub"2n+1
3

bnbn+1=class="stub"1
9
(4n2+8n+3)

①当n为偶数时:
b2=class="stub"5
3
bn=class="stub"2n+1
3

∴Wn=b1b2-b2b3+b3b4-b4b5+…+bn-1bn-bnbn+1
=b2(b1-b3)+b4(b3-b5)+…+bn(bn-1-bn+1)
=-2×class="stub"2
3
(b2+b4+…+bn)
=-class="stub"4
3
×[class="stub"n
4
(class="stub"5
3
+class="stub"2n+1
3
)]

=-class="stub"1
9
(2n2+6n)

②当n为奇数时:
b2=class="stub"5
3
bn-1=class="stub"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×class="stub"2
3
(b2+b4+…+bn-1)
+bnbn+1
=-class="stub"4
3
×[class="stub"n-1
4
(class="stub"5
3
+class="stub"2n-1
3
)
]+class="stub"1
9
(4n2+8n+3)

=class="stub"1
9
(2n2+6n+7)

Wn=
-class="stub"1
9
(2n2+6n),n为偶数
class="stub"1
9
(2n2+6n+7),n为奇数

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