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:
\(a_{n}=\frac{(-3)^{n}}{12}+\frac{1}{4}\)
