Let $f(n)=f(n-1)+f(n-2)$ be the Fibonacci sequence with $f(0)=0,f(1)=1$. Show that $$f(2n)= f(n+1)^2 - f(n-1)^2.$$
I have tried several different approaches to this problem. Both inducting from the rhs and the lhs, and I end up going in circles. Any help would be appreciated! Thanks!
Working from the lhs I've gotten: $$f(2(n+1)+1)=f(2n+1)+f(2n)$$ $$=f(2n-1) + f(2n) + f(2n)$$ $$=f(2n-1)+2(f(n+1)^2-f(n-1)^2)$$ But I can't figure out where to go from here. Help?
I found a lot of easier to prove the formula using the matrix representation for the Fibonacci numbers:
$$ \begin{pmatrix} 1 && 1 \\ 1 && 0\end{pmatrix}^n = \begin{pmatrix} F_{n+1} && F_n \\ F_n && F_{n-1}\end{pmatrix}. $$
Now, matrix multiplication fulfills $A^nA^m=A^{n+m}$, so just calculate the matrix entry $[A^{n+m}]_{1,2}$ and then set $m=n$. The result then follows.
Let's try another proof.
We know that $$F_{n+1}=F_{n-1}+F_n\tag{1}$$
There is a useful identity for the Fibonacci sequence. You can look up how it is proved here. $$F_{n+m}=F_{n-1}F_m+F_n F_{m+1}\tag{2}$$ Let's set $n=m$, then we get $$F_{2n}=F_nF_{n-1}+F_nF_{n+1}$$ Using (1), we can expand the second expression - $$F_{2n}=F_nF_{n-1}+F_n(F_n+F_{n-1})$$ $$F_{2n}=2F_nF_{n-1}+F_n^2\tag{3}$$
Now using (1), we try to find a value for the Right Hand side : $$F_{n+1}^2= (F_n+F_{n-1})^2 $$$$F_{n+1}^2 = F_n^2+F_{n-1}^2+2F_nF_{n-1}$$ $$F_{n+1}^2 - F_{n-1}^2=F_n^2+2F_nF_{n-1}\tag{4}$$
Clearly, (3) and (4) are equal. Hence proved
Let's look for the solutions to the recursive equation in the form $f(n)=C\lambda^n$. If we substitute this form into the recursion we obtain $$\lambda^n=\lambda^{n-1}+\lambda^{n-2},\tag{1}$$ which results in the quadratic equation $$1=\frac{1}{\lambda}+\frac{1}{\lambda^2}$$ whose solutions are $$\lambda_{1,2}=\frac{1\pm\sqrt{5}}{2}$$ Verify that if $C_1\lambda_1^n$ and $C_2\lambda_2^n$ are solutions of the recursive equation than their sum is a solution as well: $$f(n)=C_1\lambda_1^n+C_2\lambda_2^n,$$
where $C_1=-C_2=\sqrt{5}$ is obtained from the fact that $f(0)=1$ and $f(1)=1$ (Use eq. (1))
Now we get $$f(n+1)^2=C_1^2\lambda_1^{2n+2}+C_2^2\lambda_2^{2n+2}+2C_1C_2(\lambda_1\lambda_2)^{n+1}\tag{2}$$ $$f(n+1)^2=C_1^2\lambda_1^{2n-2}+C_2^2\lambda_2^{2n-2}+2C_1C_2(\lambda_1\lambda_2)^{n-1}\tag{3}$$ Subtract (2) and (3) keeping in mind that $\lambda_1\lambda_2=-1$ (from the quadratic equation). We arrive to $$f(n+1)^2-f(n-1)^2=C_1^2\lambda_1^{2n}\left(\lambda_1^2-\frac{1}{\lambda_1^2}\right)+C_2^2\lambda_2^{2n}\left(\lambda_2^2-\frac{1}{\lambda_2^2}\right)$$ Using the values for $C_1$,$C_2$,$\lambda_1$,and $\lambda_2$ we observe that $$C_1\lambda_1^{2n}\left(\lambda_1^2-\frac{1}{\lambda_1^2}\right)=C_2\lambda_2^{2n}\left(\lambda_2^2-\frac{1}{\lambda_2^2}\right)=1$$ and therefore $$f(n+1)^2-f(n-1)^2=C_1\lambda_1^{2n}+C_2\lambda_2^{2n}=f(2n).$$
Following up on Joseph's suggestion, we can prove the identity directly using Binet's formula $$f(n) = \frac{\varphi^n - \psi^n}{\sqrt{5}}$$ where $$\varphi = \frac{1 + \sqrt{5}}{2} \qquad \psi = \frac{1 - \sqrt{5}}{2} = -\frac{1}{\varphi}$$ so that $$\varphi^2 - \frac{1}{\varphi^2} = \sqrt{5} = \frac{1}{\psi^2} - \psi^2 \qquad \varphi\psi - \frac{1}{\varphi\psi} = 0.$$ Then \begin{align} f(n+1)^2 - f(n-1)^2 & = \left(\frac{\varphi^{n+1} - \psi^{n+1}}{\sqrt{5}}\right)^2 - \left(\frac{\varphi^{n-1} - \psi^{n-1}}{\sqrt{5}}\right)^2\\ & = \frac{1}{5}\left(\varphi^{2n+2} - 2\varphi^{n+1}\psi^{n+1} + \psi^{2n+2} - \varphi^{2n-2} + 2\varphi^{n-1}\psi^{n-1} - \psi^{2n-2}\right)\\ & = \frac{1}{5}\left(\left(\varphi^2 - \frac{1}{\varphi^2}\right)\varphi^{2n} - 2\left(\varphi\psi - \frac{1}{\varphi\psi}\right)\varphi^n\psi^n - \left(\frac{1}{\psi^2} - \psi^2\right) \psi^{2n}\right)\\ & = \frac{1}{5}\left(\sqrt{5}\varphi^{2n} - 2 \cdot 0 \cdot \varphi^n\psi^n - \sqrt{5}\psi^{2n}\right)\\ & = \frac{\varphi^{2n} - \psi^{2n}}{\sqrt{5}}\\ & = f(2n). \end{align}
Have you tried using Binet's Formula? http://mathworld.wolfram.com/BinetsFibonacciNumberFormula.html
