Question

已知函数f(x)满足：f(a+b)=f(a)·f(b),f(1)=2,则[f^2(1)+f(2)]/f(1)+[f^2(2)+f(4)]/f(3)+

[f^2(3)+f(6)]/f(5)+[f^2(4)+f(8)]/f(7)＝？

Answer 1

[f²(1)+f(2)]/f(1)+[f²(2)+f(4)]/f(3)+[f²(3)+f(6)]/f(5)+[f²(4)+f(8)]/f(7)
=[f(1+1)+f(2)]/f(1)+[f(2+2)+f(4)]/f(3)+[f(3+3)+f(6)]/f(5)+[f(4+4)+f(8)]/f(7)
=2f(2)/f(1)+2f(4)/f(3)+2f(6)/f(5)+2f(8)/f(7)
=2f(1+1)/f(1)+2f(1+3)/f(3)+2f(1+5)/f(5)+2f(1+7)/f(7)
=2f(1)f(1)/f(1)+2f(1)f(3)/f(3)+2f(1)f(5)/f(5)+2f(1)f(7)/f(7)
=2f(1)+2f(1)+2f(1)+2f(1)
=8f(1)
=16

Answer 2

f(2)=f(1)f(1)=4=2^2
f(3)=f(2)f(1)=2^3
同样推得f(x)=2^x
[f²(1)+f(2)]/f(1)+[f²(2)+f(4)]/f(3)+[f²(3)+f(6)]/f(5)+[f²(4)+f(8)]/f(7)
=[2^2+2^2]/2 + [2^4+2^4]/2^3 + [2^6+2^6]/2^5 + [2^8+2^8]/2^7
=2^2+2^2+2^2+2^2
=16

Answer 3

呃、未必也太复杂了吧！晕了！