题目内容
设f(x)=
，利用倒序相加法（课本中推导等差数列前n项和的方法），可求得f(
)+f(
)+f(
)+…f(
)的值为 ．
| 4x |
| 4x+2 |
| 1 |
| 2015 |
| 2 |
| 2015 |
| 3 |
| 2015 |
| 2014 |
| 2015 |
∵f(x)=
，
∴f（x）+f（1-x）=
+
=
+
=
+
=
+
=
=1
故可得f(
)+f(
)+f(
)+…f(
)
=f(
)+f(
)+f(
)+f(
)+…+f(
)+f(
)
=1007×1=1007
| 4x |
| 4x+2 |
∴f（x）+f（1-x）=
| 4x |
| 4x+2 |
| 41-x |
| 41-x+2 |
=
| 4x |
| 4x+2 |
| 41-x•4x |
| (41-x+2)•4x |
=
| 4x |
| 4x+2 |
| 4 |
| 4+2•4x |
| 4x |
| 4x+2 |
| 2 |
| 2+4x |
=
| 4x+2 |
| 4x+2 |
故可得f(
| 1 |
| 2015 |
| 2 |
| 2015 |
| 3 |
| 2015 |
| 2014 |
| 2015 |
=f(
| 1 |
| 2015 |
| 2014 |
| 2015 |
| 2 |
| 2015 |
| 2013 |
| 2015 |
| 1002 |
| 2015 |
| 1003 |
| 2015 |
=1007×1=1007
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