Answer:
See proof below
Step-by-step explanation:
We have to verify that if we substitute
in the equation
the equality is true.
Let's substitute first in the right hand side:
![3a_{n-1}+4a_{n-2}=3(4^{n-1}+2(-1)^{n-1})+4(4^{n-2}+2(-1)^{n-2})](https://tex.z-dn.net/?f=3a_%7Bn-1%7D%2B4a_%7Bn-2%7D%3D3%284%5E%7Bn-1%7D%2B2%28-1%29%5E%7Bn-1%7D%29%2B4%284%5E%7Bn-2%7D%2B2%28-1%29%5E%7Bn-2%7D%29)
Now we use the distributive laws. Also, note that
(this also works when the power is n-2).
![=3(4^{n-1})+6(-1)^{n-1}+4(4^{n-2})+8(-1)^{n-2}](https://tex.z-dn.net/?f=%3D3%284%5E%7Bn-1%7D%29%2B6%28-1%29%5E%7Bn-1%7D%2B4%284%5E%7Bn-2%7D%29%2B8%28-1%29%5E%7Bn-2%7D)
![=3(4^{n-1})+(-1)(6)(-1)^{n}+4^{n-1}+(-1)^2(8)(-1)^{n}](https://tex.z-dn.net/?f=%3D3%284%5E%7Bn-1%7D%29%2B%28-1%29%286%29%28-1%29%5E%7Bn%7D%2B4%5E%7Bn-1%7D%2B%28-1%29%5E2%288%29%28-1%29%5E%7Bn%7D)
![=4(4^{n-1})-6(-1)^{n}+8(-1)^{n}=4^n+2(-1)^n=a_n](https://tex.z-dn.net/?f=%3D4%284%5E%7Bn-1%7D%29-6%28-1%29%5E%7Bn%7D%2B8%28-1%29%5E%7Bn%7D%3D4%5En%2B2%28-1%29%5En%3Da_n)
then the sequence solves the recurrence relation.
Answer:
It would be 55.88 but if you round it it would be 55.9 then round it again it would be 56%.
3 remainder of 7 i fink
Step-by-step explanation: