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My question on it is: No matter what the method I use, the answers must be the same, right? method 1) directly substitute f(n-1), f(n-2) to find f(n), we get d is the final answer.
for the first question.
Well, clearly if answer is not same, then you are incorrectly using method. :P

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How about this: f(n) = -4f(n-1) -3f(n-2) We can go backward like f(n+2) = -4 f(n+1) -3f(n)
that give us the characteristic equation for recursive formula is: r^2 +4r+3 =0, \(r_1= -3 \) \(r_2= -1\) Hence the general solution for it is \(f(n) = C_1 (-3)^n +C_2(-1)^n\)
f(0) = -5, hence -5 = C1 + C2 f(1) = 20 , hece 20 = -3C1-C2 solve them, it gives me C1 = -15/2 C2 = 5/2
yes d is correct .... .-. \[\rm f(2) = -4•f(2 -1) - 3•f(2 - 2) \]\[\rm f(2) = -4•f(1) - 3•f(0) \] =-65 :-)???
@Nnesha Loser66 is more of looking for a way to find explicit formula.
otay.
oh, I know my mistake. hihihi. it works well just the way I count f(3) is f(2) in the sequence. hehehe.. Thanks you all.
@Nnesha my goal is to apply my knowledge in Discrete Math to put it in logic
otay. http://prntscr.com/7fakty
\(f(n) = \dfrac{5}{2}(-1)^n-\dfrac{15}{2} (-3)^n\) f(2) , that is n = 2 , \(f(2) = \dfrac{5}{2} -\dfrac{15}{2}*9= 65\) f(3) , that is n =3 , \(f(3) = \dfrac{5}{2}(-1)^3 -\dfrac{15}{2}(-3)^3=200\)
oh, f(2) = -65 :)
the first term is f(0) , next is f(1),
That was my mistake. hehehe...
gO_OD job! @Loser66 ;-)
haaaaaaaaaaaaaahahaha... thank you for the tough flower. @Nnesha
flower or chocolates ? Yw

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