We give two applications of the formula for sequences satisfying second order recurrence relations:

1. Recall that the Fibonacci sequence satisfies the recurrence relation $$ f_{n+1}=f_n + f_{n-1}. $$ Thus, $f_0=1$ $A=1$ and $B=1$ Therefore, the theorem yields the following formula for the Fibonacci sequence: $$ f_n=\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor} \binom{n-k}{k} $$
2. Fix a prime $p$ and define a sequence $s$ by $s_n=\tau(p^n)$ where $\tau$ denotes the Ramanujan tau function. Recall that $\tau$ satisfies $$ \tau(p^{n+1})=\tau(p)\tau(p^n)-p^{11}\tau(p^{n-1}). $$ Thus, $s_0=1$ $A=\tau(p)$ and $B=-p^{11}$ Therefore, the theorem yields $$ \tau(p^n)=\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor} \binom{n-k}{k} (-p^{11})^k (\tau(p))^{n-2k}. $$ This formula is valid for all primes $p$ and all nonnegative integers $n$