Resposta :
Olá, Alfredo.
[tex]f(x + y) = f(x) . f( y )\ e\ f( 1 ) = 2\\\\ f(1+1)=f(2)=f(1).f(1)=2.2=4 \Rightarrow f(2)=4\\\\ f(1+2)=f(3)=f(1).f(2)=2.4=8 \Rightarrow f(3)=8\\\\ f(2+2)=f(4)=f(2).f(2)=4.4=16 \Rightarrow f(4)=16\\\\ f(3+2)=f(5)=f(3).f(2)=8.4=32 \Rightarrow f(5)=32\\\\ f(3+3)=f(6)=f(3).f(3)=8.8=64 \Rightarrow f(6)=64[/tex]
[tex]\Rightarrow f(1),...,f(6)[/tex] formam uma PG de razão [tex]q=2[/tex] e [tex]a_1=2[/tex] .
A soma [tex]f(1)+...+f(6)[/tex] é, portanto, a soma desta PG.
[tex]\therefore f(1)+...+f(6)=\frac{a_1(q^n-1)}{q-1}=\frac{2.(2^6-1)}{2-1}=2\cdot(64-1)=2\cdot63=\\\\ =126[/tex]