Answer:
See proof below
Step-by-step explanation:
We have to verify that if we substitute [tex]a_n=4^n+2(-1)^n[/tex] in the equation [tex]a_n=3a_{n-1}+4a_{n-2}[/tex] the equality is true.
Let's substitute first in the right hand side:
[tex]3a_{n-1}+4a_{n-2}=3(4^{n-1}+2(-1)^{n-1})+4(4^{n-2}+2(-1)^{n-2})[/tex]
Now we use the distributive laws. Also, note that [tex](-1)^{n-1}=\frac{1}{-1}(-1)^n=(-1)(-1)^{n}[/tex] (this also works when the power is n-2).
[tex]=3(4^{n-1})+6(-1)^{n-1}+4(4^{n-2})+8(-1)^{n-2}[/tex]
[tex]=3(4^{n-1})+(-1)(6)(-1)^{n}+4^{n-1}+(-1)^2(8)(-1)^{n}[/tex]
[tex]=4(4^{n-1})-6(-1)^{n}+8(-1)^{n}=4^n+2(-1)^n=a_n[/tex]
then the sequence solves the recurrence relation.
