How to find $\Delta a_n$? - recurrence relation

Question

Sequences fulfills recurrence $a_{n+2}=a_{n+1}-2a_n$ and $ 2a_{n+2}=3a_{n+1}+2a_n$.

Find $\Delta a_n$ if $a_0=1$

Answer 1

Sequences fulfills recurrence $a_{n+2}=a_{n+1}-2a_n$ and $ 2a_{n+2}=3a_{n+1}+2a_n$.

Find $\Delta a_n$ if $a_0=1$

$$2(a_{n+1}-2a_n)=3a_{n+1}+2{a_n}$$ $$0=a_{n+1}+6a_n$$ $$a_{n+1}=-6a_n$$ $$a_{n+2}=36a_n$$

$$36a_n=-6a_n-2a_n=-8a_n$$

$36a_n=-8a_n$ only when $a_n=0$