Prove recurrence relation using mathematical induction

Question

solve recurrence relation $a_n = a_{n–1} + 12 a_{n–2}$, where $a_0 = 1$ and $a_1 = 11$ and Verify, using Principle of Mathematical Induction, that $a_n = (-1)(-3)^n + (2)(4^n)$.

ans: i have done so far.. put $a_n=b_n$
$a_n-a_{n-1}-12a_{n-2}=0$
$b_n-b_{n-2}-12a_{n-2}=0$
$b^2-b-12=0$, $b=-3,4$

Answer 1

$a_{n+1} = a_n + 12a_{n-1}=$

$[(-1)(-3)^n + (2)(4^n)] + 12[(-1)(-3)^{n-1} + (2)(4^{n-1})]=$

.....

$(-1)(-3)^{n+1} + (2)(4^{n+1})$

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