题目内容
计算:(1)若数列an=
,求
(a2+a3+a4+…+an);
(2)若函数f(x)=
在R上是连续函数,求a的取值.
| 1 |
| n(n-1) |
| lim |
| n→∞ |
(2)若函数f(x)=
|
(1)∵an=
=
-
,
a2+a3+a4+…+an
=(1-
)+(
-
)+(
-
)+…+(
-
)
=1-
.
∴
(a2+a3+a4+…+an)
=
(1-
)
=1.
(2)∵函数f(x)=
,
∴
f(x)=
(a+2x)=a+2,
f(x)=
=
=
=
.
∵f(x)在R上是连续函数,
∴a+2=
,
∴a=-
.
| 1 |
| n(n-1) |
| 1 |
| n-1 |
| 1 |
| n |
a2+a3+a4+…+an
=(1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| n-1 |
| 1 |
| n |
=1-
| 1 |
| n |
∴
| lim |
| n→∞ |
=
| lim |
| n→∞ |
| 1 |
| n |
=1.
(2)∵函数f(x)=
|
∴
| lim |
| x→1- |
| lim |
| x→1- |
| lim |
| x→1+ |
| lim |
| x→1+ |
| ||
| x(x-1) |
=
| lim |
| x→1+ |
| ||||
x(
|
=
| lim |
| x→1+ |
| 1 | ||
x(
|
=
| 1 |
| 2 |
∵f(x)在R上是连续函数,
∴a+2=
| 1 |
| 2 |
∴a=-
| 3 |
| 2 |
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