Question

1/3,0,1,-2 what are the next 3 terms of the sequence

Answer 1

Hello,

One solution may be
$a_{n+1}=a_{n}*(-3) +1$

$a_{0}=\frac{1}{3}$

$a_{1}=\frac{1}{3}*(-3) +1=0$
$a_{2}=0*(-3) +1=1$
$a_{3}=1*(-3) +1=-2$

$a_{4}=(-2)*(-3) +1=7$
$a_{5}=7*(-3) +1=-20$
$a_{6}=(-20)*(-3) +1=61$
...
REM:
$a_{n}=\frac{(-3)^{n}}{12}+\frac{1}{4}$