Solve the initial value problem for the sequence $\left \{ u_{n}| n \in \mathbb{N} \right \}$ satisfying the recurrence relation: $u_n − 5u_{n-1} + 6u_{n−2} = 0 $ with $u_0 = 1$ and $u_1 = 1$.

Ive gotten the general solution to be $u_n = A(2)^n + B(3)^n$.

Once I sub the initial values:

$u_0 = 2A + 3B = 1$

$u_1 = 2A + 3B = 1$

And Im unsure on how to solve this system. Any help appreciated, thanks.

  • 1
    $\begingroup$ Plug in $n=0$ to get $u_0 = A+B$. Yours is wrong. $\endgroup$ – GEdgar May 29 '17 at 10:34
  • $\begingroup$ @GEdgar sorry, what did i do wrong? $\endgroup$ – Rob Bor May 29 '17 at 10:37
  • $\begingroup$ we get $$u_n=2^{n+1}-3^n$$ $\endgroup$ – Dr. Sonnhard Graubner May 29 '17 at 10:45
  • $\begingroup$ @Dr.SonnhardGraubner How? $\endgroup$ – Rob Bor May 29 '17 at 10:50
  • $\begingroup$ $2^0 = 1, 3^0 = 1$. $\endgroup$ – GEdgar May 29 '17 at 12:27

$$ u_0 = A 2^0 + B 3^0 = A + B \\ u_1 = A 2^1 + B 3^1 = 2A + 3B $$ So you got the first equation wrong.


These problems of generalized Fibonacci sequences seem to show up here weekly. Here, then, is a generalized solution for solving all such problems with a recursion formula of the type $f_n=af_{n-1}+bf_{n-2}$ and arbitrary initial conditions.

There have been many extensions of the sequence with adjustable (integer) coefficients and different (integer) initial conditions, e.g., $f_n=af_{n-1}+bf_{n-2}$. (You can look up Pell, Jacobsthal, Lucas, Pell-Lucas, and Jacobsthal-Lucas sequences.) Maynard has extended the analysis to $a,b\in\mathbb{R}$, (Ref: Maynard, P. (2008), “Generalised Binet Formulae,” $Applied \ Probability \ Trust$; available at http://ms.appliedprobability.org/data/files/Articles%2040/40-3-2.pdf.)

We have extended Maynard's analysis to include arbitrary $f_0,f_1\in\mathbb{R}$. It is relatively straightforward to show that

$$f_n=\left(f_1-\frac{af_0}{2}\right) \frac{\alpha^n-\beta^n}{\alpha-\beta}+\frac{af_0}{2} \frac{\alpha^n+\beta^n}{\alpha+\beta}= \left(f_1-\frac{af_0}{2}\right)F_n+\frac{af_0}{2}L_n$$

where $\alpha,\beta=(a\pm\sqrt{a^2+4b})/2$, $F_n=\frac{\alpha^n-\beta^n}{\alpha-\beta}$, and $L_n=\frac{\alpha^n+\beta^n}{\alpha+\beta}$.

The result is written in this form to underscore that it is the sum of a Fibonacci-type and Lucas-type Binet-like terms. It will also reduce to the standard Fibonacci and Lucas sequences for $a=b=1, f_1=1, \text{ and } f_0=0 \text{ or }2$.

This can also be expressed as



$$u_n-2u_{n-1}=3(u_{n-1}-2u_{n-2}),$$ which says that $u_{n}-2u_{n-1}=-3^{n-1}$.

Thus, $$u_n-2u_{n-1}=-3^{n-1}$$ $$2^1u_{n-1}-2^2u_{n-2}=-2^13^{n-2}...$$ $$2^{n-1}u_1-2^nu_0=-2^{n-1}3^0,$$ which after summing gives: $$u_n-2^nu_0=-3^{n-1}-2\cdot3^{n-2}-...-2^{n-1}$$ or $$u_n=2^n-\frac{3^n-2^n}{3-2}$$ or $$u_n=2^{n+1}-3^n.$$ Done!


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