Se f (1) = 160 ef (n + 1) = –2f (n), o que é f (4)?
Responda:
#f(4)=-1280#
Explicação:
As #f(1)=160# e #f(n+1)=–2f(n)#
#f(2)=-2f(1)=-2xx160=-320#
#f(3)=-2f(2)=-2xx(-320)=640#
#f(4)=-2f(3)=-2xx640=-1280#
#f(4)=-1280#
As #f(1)=160# e #f(n+1)=–2f(n)#
#f(2)=-2f(1)=-2xx160=-320#
#f(3)=-2f(2)=-2xx(-320)=640#
#f(4)=-2f(3)=-2xx640=-1280#