Question

Please help meee
Solve the 2nd order non-homogeneous linear recurrence relations with constant coefficients: f(n+2)-6f(n+1)+9f(n)=36 x 3^n subject to the start-up conditions f(0)=2; f(1)=27

Answer 1

I once could do this...

Answer 2

...u once could do everything lol

Answer 3

I think.

Answer 4

\[f(n+2)-6f(n+1)+9f(n)=36\cdot 3^n\quad;\quad f(0)=2,f(1)=27\]\[\lambda^2-6\lambda+9=(\lambda-3)^2=0\quad\Rightarrow\quad\lambda_{1,2}=3\]\[f_1(n)=(C_1+C_2n)\cdot 3^n\]\[f_2(n)=an^2\cdot 3^n\]\[f(n+2)-6f(n+1)+9f(n)=\]\[=a(n+2)^2\cdot 3^{n+2}-6\cdot a(n+1)^2\cdot 3^{n+1}+9\cdot an^2\cdot 3^n=\]\[=a\left(9(n+2)^2-18(n+1)^2+9n^2\right)\cdot 3^n=18a\cdot 3^n=36\cdot 3^n\quad\Rightarrow\quad a=2\]\[f_2(n)=2n^2\cdot 3^n\]\[f(n)=(C_1+C_2n+2n^2)\cdot 3^n\]\[f(0)=C_1=2\]\[f(1)=(C_1+C_2+2)\cdot 3=27\quad\Rightarrow\quad C_2=5\]\[f(n)=(2+5n+2n^2)\cdot 3^n=(2+n)(1+2n)\cdot 3^n\]

Answer 5

nik!!!!!!!!! thank you sooo much!! u are the best =]

Answer 6

hey nik are u still there?

Answer 7

yes

Answer 8

hey do you think you could help me with solving couple of fibonacci problems please?

Answer 9

where are problems?

Answer 10

In these problems \[f _{n}\] denotes the nth Fibonacci number

1) Prove \[\sum_{k=0}^{n} f _{k}^{2} = f _{n} f _{n+1}\]

2) Prove \[f _{n+1} f _{n-1} -f _{n}^2 = (-1)^n\]

Answer 11

3) Let n be an even (positive) integer. What is \[\sum_{k=0}^{n/2}\left(\begin{matrix}n-k \\ k\end{matrix}\right)?\]
(Hint: Calculate this explicitly for n=2,4,6,8... and then GUESS). No proof required..just the right guess