How can you prove this by strong induction? The sequence $b_1,b_2,...$ is defined recursively as:\begin{align} b_1&=0;\\ b_2&=1;\\ b_n&=2b_{n-1}-2b_{n-2}-1 \ \text{for} \ n\geq3. \end{align} Prove that this means: $$\forall n\geq1: b_n=(\sqrt{2})^n \sin{\left(\frac{1}{4}\pi n \right)}-1$$
Edit:
I have tried to prove this by strong induction and have verified than $P(1)$ and $P(2)$ is true, where $P(n)$ is the statement $b_n=(\sqrt{2})^n \sin{\left(\frac{1}{4}\pi n \right)}-1$. 
edit: I have assumed that $P(k-2)$ and $P(k-1)$ is true for $k\in\mathbb{N}$. Then I have managed to simplify to: $$b_n=(\sqrt{2})^n \left[\sqrt{2}\sin{\left(\frac{1}{4}\pi (n-1) \right)}- \sin{\left(\frac{1}{4}\pi (n-2) \right)}\right]-5$$
I can't simplify any further. 

Since you have verified the base case, we proceed to the inductive case. Fix $n >2$ and suppose that for all $k <n$, $P(k)$ is true. We want to show that $P(n)$ is true. 
$$
\begin{aligned}
b_{n}&=2b_{n-1}-2b_{n-2}-1\\
&=2\left((\sqrt{2})^{n-1}\sin\left(\frac{1}{4}\pi (n-1)\right)-1\right)-2\left((\sqrt{2})^{n-2}\sin\left(\frac{1}{4}\pi (n-2)\right)-1\right)-1\\
&=(\sqrt{2})^{n}\left(\sqrt{2}\sin\left(\frac{\pi (n-1)}{4}\right)-\sin\left(\frac{\pi (n-2)}{4}\right)\right) - 1
\end{aligned}
$$
Now, using $\sin(A+B)=\sin A\cos B+\cos A\sin B$, we get
$$
\begin{aligned}
&\sqrt{2}\sin\left(\frac{\pi (n-1)}{4}\right)-\sin\left(\frac{\pi (n-2)}{4}\right)=\\
&\sqrt{2}\left(\sin\left(\frac{\pi n}{4}\right)\cos\left(\frac{-\pi}{4}\right)-\cos\left(\frac{\pi n}{4}\right)\sin\left(\frac{-\pi}{4}\right)\right)-\sin\left(\frac{\pi n}{4}\right)\cos\left(\frac{-\pi}{2}\right)+\cos\left(\frac{\pi n}{4}\right)\sin\left(\frac{-\pi}{2}\right)\\
&=\sqrt{2}\left(\sin\left(\frac{\pi n}{4}\right)\frac{1}{\sqrt{2}}-\cos\left(\frac{\pi n}{4}\right)(\frac{-1}{\sqrt{2}})\right)-\cos\left(\frac{\pi n}{4}\right)\\
&=\sin\left(\frac{\pi n}{4}\right)+\cos\left(\frac{\pi n}{4}\right)-\cos\left(\frac{\pi n}{4}\right)\\
&=\sin\left(\frac{\pi n}{4}\right)
\end{aligned}
$$
