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Mathematics 8 Online
OpenStudy (anonymous):

Solve the 2nd order homogeneous linear recurrence relation with constant coefficients: f(n+2)-4f(n+1)+4f(n)=0 subject to the start-up conditions f(0)=3; f(1)=4

OpenStudy (nikvist):

\[ f(n+2)-4f(n+1)+4f(n)=0\quad;\quad f(0)=3,f(1)=4\]\[\lambda^2-4\lambda+4=(\lambda-2)^2=0\quad\Rightarrow\quad\lambda_{1,2}=2\]\[f(n)=(C_1+C_2n)2^n\]\[f(0)=C_1=3\]\[f(1)=2(C_1+C_2)=4\quad,\quad C_2=-1\]\[f(n)=(3-n)2^n\]

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