Please Help!! I have no idea how to do this and this is a take home final exam and i have to pass this to pass the class... thankss

Solve the 2nd order homogeneous linear recurrence relation with constant coefficients f(n+2)-3f(n+1)+2f(n)=0 subject to the start-up conditions f(0)=5 & f(1)=8

Question

Please Help!! I have no idea how to do this and this is a take home final exam and i have to pass this to pass the class... thankss

Solve the 2nd order homogeneous linear recurrence relation with constant coefficients f(n+2)-3f(n+1)+2f(n)=0 subject to the start-up conditions f(0)=5 &  f(1)=8

anonymous · Answer

f(2)=14

anonymous · Answer

can u show me how to do this because i think he wants to see the work

mathmate · Answer

Using linear algebra or generator functions?

anonymous · Answer

i dont even know what that means..sorry im just not too good in discrete math... I typed exactly what is on the page

anonymous · Answer

put n=0
anf and replace f(0) with 5 and f(1) with 8

anonymous · Answer

*and

anonymous · Answer

what is f(0) and f(1)?

mathmate · Answer

So the course is discrete maths?
Does anything like
1/(1-x) = 1+x+x^2+x^3+... mean anything to you?

anonymous · Answer

umm im not even sure =/

anonymous · Answer

i just need to know how to do this step by step ..cause i don't have the slightest clue

nikvist · Answer

$$f(n+2)-3f(n+1)+2f(n)=0\quad;\quad f(0)=5,f(1)=8$$$$f(n)=C_1\lambda_1^n+C_2\lambda_2^n$$$$\lambda^2-3\lambda+2=(\lambda-1)(\lambda-2)=0$$$$f(n)=C_1+C_2 2^n$$$$f(0)=C_1+C_2=5$$$$f(1)=C_1+2C_2=8$$$$C_1=2,C_2=3$$$$f(n)=2+3\cdot2^n$$

anonymous · Answer

nikvist is that the complete problem solved?

nikvist · Answer

yes

anonymous · Answer

what about the f(2) = 14

anonymous · Answer

plug n=2 and f(2) is 14