题目内容
设f(x)=
,定义f(1)是代数式
当x=1时的值,即f(1)=
=
,同理f(2)=
=
,f(
)=
=
,…根据此运算,求f(1)+f(
)+f(2)+f(
)+f(3)+f(
)+f(4)+…+f(
)+f(n)=______.
| x2 |
| x2+1 |
| x2 |
| x2+1 |
| 12 |
| 12+1 |
| 1 |
| 2 |
| 22 |
| 22+1 |
| 4 |
| 5 |
| 1 |
| 2 |
(
| ||
(
|
| 1 |
| 5 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| n |
由题意可知f(3)=
=
,f(
)=
=
,f(4)=
,f(
)=
,
∴f(2)+f(
)=1,f(3)+f(
)=1,f(4)+f(
)=1,…f(n)+f(
)=1,
∴原式=
+(n-1)=n-
.故答案为:n-
.
| 32 |
| 32+1 |
| 9 |
| 10 |
| 1 |
| 3 |
(
| ||
(
|
| 1 |
| 10 |
| 16 |
| 17 |
| 1 |
| 4 |
| 1 |
| 17 |
∴f(2)+f(
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| n |
∴原式=
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
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