题目内容
设f(x)=
,利用课本中推导等差数列前n项和的公式的方法,可求得f(-3)+f(-2)+…+f(0)+…+f(3)+f(4)的值为______.
| 1 |
| 4x+2 |
设a+b=1,则f(a)+f(b)=
+
=
+
=
+
=
=
.
所以f(-3)+f(4)=
,f(-2)+f(3)=
,f(-1)+f(2)=
,f(0)+f(1)=
,
f(-3)+f(-2)+…+f(0)+…+f(3)+f(4)=4×
=2.
故答案为:2.
| 1 |
| 4a+2 |
| 1 |
| 4b+2 |
=
| 4b |
| (4a+2)4b |
| 1 |
| 4b+2 |
| 4b |
| 4+2•4b |
| 1 |
| 4b+2 |
| 4b+2 |
| 2(4b+2) |
| 1 |
| 2 |
所以f(-3)+f(4)=
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
f(-3)+f(-2)+…+f(0)+…+f(3)+f(4)=4×
| 1 |
| 2 |
故答案为:2.
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