Base case: For $n = 1$, we have $\gcd(b_1, b_2) = \gcd(3, 4) = 1$. So the base case holds.
Inductive step: Assume that $\gcd(b_n, b_{n+1}) = 1$ for some $n \geq 1$. We want to show that $\gcd(b_{n+1}, b_{n+2}) = 1$.
We know that $b_{n+2} = 2b_{n+1} + 2b_n$.
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