Inductive proof for recursive formula $a_n=4-a_{n-1}$

Question

I have a recursive formula $a_n=4-a_{n-1}, a_3=7$ and I had to solve this and check the correction of the answer with induction.

I started solving this and the solved formula is: $a_n=-2((-1)^{n-3}-1)+7(-1)^{n-3}$.

1. Base case $n=3$ is true

2. Let's assume this is true for $n=k$

3. Thus this should be true for $n=k+1$

   $a_{n=k+1}=5(-1)^{k-2}+2$.

I don't know how to continue from this place, please someone help me and explain this, thank you.

Answer 1

Note that $-2((-1)^{n-3} - 1) + 7(-1)^{n-3} = 5(-1)^{n-3} + 2$.

Assuming $a_n = 5(-1)^{n-3} + 2$, the recursive formula tells us that

$$a_{n+1} = 4 - a_n = 4 - [5(-1)^{n-3} + 2] = -5(-1)^{n-3} + 2 = 5(-1)^{n-2}+2.$$

This proves the result.

Answer 2

Hint: Prove that $a_{n+2}=a_n$ for all $n$.