Prove by induction that $b(n)=b(n-1) + b(n-2) \space\text{for}\space n\geq3$

Question

It is given that $b(1)=1 \space\text{and}\space b(n)=a(n+1)+a(n-1) \space\text{for}\space n\geq2$.

Prove by induction that $b(n)=b(n-1) + b(n-2) \space\text{for}\space n\geq3.$

Here, $a(n)$ is the Fibonacci sequence.

Answer 1

I did not really use Induction, I just use the definition of the sequence in terms of the Fibonacci numbers and in the end I show that it is also true for $n=3$ directly. $ \text{The fibonacci sequence is defined as:}\\ a(n+1)=a(n)+a(n-1)\quad \quad \\ \text{I have:}\\ b(n)=a(n+1)+a(n-1)\\ \Rightarrow b(n)=a(n)+a(n-1)+a(n-2)+a(n-3)\\ b(n)=[a(n)+a(n-2)]+[a(n-1)+a(n-3)]\\ \text{apply the definition of the sequence to get:}\\ b(n)=b(n-1)+b(n-2)\\ \text{now I just need to check that the statement is true for $n=3$}\\ b(1)=1\\ b(2)=a(3)+a(1)=3\\ b(3)=a(4)+a(2)=4\\ \Rightarrow b(3)=b(2)+b(1)$