## Finding the Rule For Simple Recurrence Relations

3,11,35,107, 323,...

The first differences are

8, 22, 72, 216...

The second differences are

14, 50, 144

Neither the first or second differences are constant so the sequence is not linear or quadratic. It could however, be a simple recurrence relationship,

\[u_{n+1} = a u_n +b\]

.If it is we can easily found out. Divide each term by the previous on. In this case we obtain

\[\frac{11}{3}=6.666..., \frac{35}{11}=3.18181818..., \frac{107}{35}=3.0571428...,\frac{323}{107}=3.02...\]

Suceesive ratios are getting closer to 3. If the rule is of the forma above, we must have

\[a=3\]

. To find b, substitute the first two erms into re relationship, giving\[11=3 \times 3 +b \rightarrow b=2\]

The rule is then

\[u_{n+1} = 3 u_n +2\]

and it is easy to see that this generates the sequence.