Determine the closed form of an=2an−1−2an−2a_n = 2a_{n-1} - 2a_{n-2}an=2an−1−2an−2 with a0=1,a1=1a_0=1, a_1=1a0=1,a1=1.
an=(2)n(cos(nπ4)+sin(nπ4))a_n = (\sqrt{2})^n (\cos(\frac{n\pi}{4}) + \sin(\frac{n\pi}{4}))an=(2)n(cos(4nπ)+sin(4nπ))
an=2ncos(nπ4)a_n = 2^n \cos(\frac{n\pi}{4})an=2ncos(4nπ)
an=(2)ncos(nπ4)a_n = (\sqrt{2})^n \cos(\frac{n\pi}{4})an=(2)ncos(4nπ)
an=2n(cos(nπ4)+sin(nπ4))a_n = 2^n (\cos(\frac{n\pi}{4}) + \sin(\frac{n\pi}{4}))an=2n(cos(4nπ)+sin(4nπ))