题目内容
设f(x)=
.则f(
)+f(
)+f(
)+…+f(
)______.
| 4x |
| 4x+2 |
| 1 |
| 2013 |
| 2 |
| 2013 |
| 3 |
| 2013 |
| 2012 |
| 2013 |
若a+b=1,则f(a)+f(b)=
+
=
=
=
=1,
所以f(
)+f(
)+f(
)+…+f(
)
=[f(
)+f(
)]+[f(
)+f(
)]+…+[f(
)+f(
)]
=1+1+…+1=1006.
故答案为:1006.
| 4a |
| 4a+2 |
| 4b |
| 4b+2 |
=
| 4a(4b+2)+4b(4a+2) |
| (4a+2)(4b+2) |
=
| 2?4a+b+2(4a+4b) |
| 4a+b+2(4a+4b)+4 |
=
| 8+2(4a+4b) |
| 8+2(4a+4b) |
所以f(
| 1 |
| 2013 |
| 2 |
| 2013 |
| 3 |
| 2013 |
| 2012 |
| 2013 |
=[f(
| 1 |
| 2013 |
| 2012 |
| 2013 |
| 2 |
| 2013 |
| 2011 |
| 2013 |
| 1006 |
| 2013 |
| 1007 |
| 2013 |
=1+1+…+1=1006.
故答案为:1006.
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