问题
填空题
已知函数f(x)满足f(m+n)=f(m)f(n),f(1)=4,则
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答案
∵函数f(x)满足f(m+n)=f(m)f(n),
∴令m=n,得f(2n)=f(n)f(n),即f(2n)=f2(n),
因此f(2)=f2(1),f(4)=f2(2),f(6)=f2(3),f(8)=f2(4),f(10)=f2(5),
∴
+f2(1)+f(2) f(1)
+f2(2)+f(4) f(3)
+f2(3)+f (6) f(5)
+f2(4)+f(8) f(7) f2(5)+f(10) f(9)
=
+2f2(1) f(1)
+2f2(2) f(3)
+2f2(3) f(5)
+2f2(4) f(7) 2f2(5) f(9)
又∵f2(n)=f(n)f(n)=f(n+n)=f(2n-1+1)=f(2n-1)•f(1)
∴
=f(1),可得f2(n) f(2n-1)
=2f2(1) f(1)
=2f2(2) f(3)
=2f2(3) f(5)
=2f2(4) f(7)
=2f(1)=8,2f2(5) f(9)
因此,
+f2(1)+f(2) f(1)
+f2(2)+f(4) f(3)
+f2(3)+f (6) f(5)
+f2(4)+f(8) f(7)
=5×8=40f2(5)+f(10) f(9)
故答案为:40