题目内容
如果函数f(x)满足f(a+b)=f(a)•f(b)且f(1)=2则
+
+
+
+…+
+
=______.
| f(2) |
| f(1) |
| f(3) |
| f(2) |
| f(4) |
| f(3) |
| f(6) |
| f(5) |
| f(2010) |
| f(2009) |
| f(2012) |
| f(2011) |
∵函数f(x)满足f(a+b)=f(a)•f(b)
∴f(n+1)=f(n)f(1)
∵f(1)=2
∴
=f(1)=2
∴
=f(1),
=f(1),
=f(1)…
=f(1)
∴
+
+
+
+…+
+
=2011f(1)=2011×2=4022
故答案为:4022
∴f(n+1)=f(n)f(1)
∵f(1)=2
∴
| f(n+1) |
| f(n) |
∴
| f(2) |
| f(1) |
| f(3) |
| f(2) |
| f(4) |
| f(3) |
| f(2012) |
| f(2011) |
∴
| f(2) |
| f(1) |
| f(3) |
| f(2) |
| f(4) |
| f(3) |
| f(6) |
| f(5) |
| f(2010) |
| f(2009) |
| f(2012) |
| f(2011) |
故答案为:4022
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