计算:(1)若数列an=1n(n-1),求limn→∞(a2+a3+a4+…+an);(2)若函数f(x)=x-1x•(x-1)(x>1)a+2x(x≤1)在R上是连续函数,求a的取值.-数学

题目简介

计算:(1)若数列an=1n(n-1),求limn→∞(a2+a3+a4+…+an);(2)若函数f(x)=x-1x•(x-1)(x>1)a+2x(x≤1)在R上是连续函数,求a的取值.-数学

题目详情

计算:(1)若数列an=
1
n(n-1)
,求
lim
n→∞
(a2+a3+a4+…+an)

(2)若函数f(x)=
x
-1
x•(x-1)
(x>1)
a+2x(x≤1)
在R上是连续函数,求a的取值.
题型:解答题难度:中档来源:不详

答案

(1)∵an=class="stub"1
n(n-1)
=class="stub"1
n-1
-class="stub"1
n

a2+a3+a4+…+an
=(1-class="stub"1
2
)+(class="stub"1
2
-class="stub"1
3
)+(class="stub"1
3
-class="stub"1
4
)+…+(class="stub"1
n-1
-class="stub"1
n
)

=1-class="stub"1
n

lim
n→∞
(a2+a3+a4+…+an)

=
lim
n→∞
(1-class="stub"1
n
)

=1.
(2)∵函数f(x)=
x
-1
x•(x-1)
(x>1)
a+2x(x≤1)

lim
x→1-
f(x)
=
lim
x→1-
 (a+2x)
=a+2,
lim
x→1+
f(x)
=
lim
x→1+
x
-1
x(x-1)

=
lim
x→1+
x
-1
x(
x
-1)(
x
+1)

=
lim
x→1+
class="stub"1
x(
x
+1)

=class="stub"1
2

∵f(x)在R上是连续函数,
∴a+2=class="stub"1
2

∴a=-class="stub"3
2

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