Question

Prove that an = 4^n + 2(-1)^nis the solution to

Answer 1

Answer:

See proof below

Step-by-step explanation:

We have to verify that if we substitute $a_n=4^n+2(-1)^n$ in the equation $a_n=3a_{n-1}+4a_{n-2}$ the equality is true.

Let's substitute first in the right hand side:

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

Now we use the distributive laws. Also, note that $(-1)^{n-1}=\frac{1}{-1}(-1)^n=(-1)(-1)^{n}$ (this also works when the power is n-2).

$=3(4^{n-1})+6(-1)^{n-1}+4(4^{n-2})+8(-1)^{n-2}$

$=3(4^{n-1})+(-1)(6)(-1)^{n}+4^{n-1}+(-1)^2(8)(-1)^{n}$

$=4(4^{n-1})-6(-1)^{n}+8(-1)^{n}=4^n+2(-1)^n=a_n$

then the sequence solves the recurrence relation.