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Can someone please help me understand this? This is what I got but it isn't correct.

$a_n=14a_{n−1}−33a_{n−2}$ for $n≥3$ with initial conditions $a_0=−24,a_1=−200$.

Solve for $a_n$.

$t^2 = -14t+33$

$r_1=-11$

$r_2= -3$

$n_0 = M(-11)^0 + N(-3)^0 = -24$

$n_1 = M(-11)^1 + N(-3)^1 = -200$

By algebra, $m = 34 , n = -58$, so

$a_n= 34(-11)^n - 58(-3)^n$.

Ken
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user3395349
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    The equation should be $t^{2} - 14 t + 33 = 0$ which has solutions $t \in { 11, 3 }$. From this $a_{n} = A \cdot 11^{n} + B \cdot 3^{n}$. For the initial conditions the values $A = - 16$ and $B = -8$ are found. Thus, $a_{n} = - 8 [ 2 \cdot 11^{n} + 3^{n}]$ – Leucippus Jun 09 '15 at 22:54
  • Thank you so much for your help, where I am confused is how you got {11,3} instead of {-11,-3} wouldnt (t-3)(t-11) = t^2 -14t +33 ? – user3395349 Jun 09 '15 at 23:12

2 Answers2

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Just a sign error in the setup. You suppose that $a_n = t^n$ is the general rule, and plugging into the recurrence relation (for $n \geq 3$) gives

$$t^n = 14 t^{n-1} - 33 t^{n-2},$$

or, putting everything on one side,

\begin{align} t^n - 14 t^{n-1} + 33 t^{n-2} &= 0\\ t^{n-2}(t^2 - 14 t + 33) & = 0\\ \end{align}

[I trust the nature of the sign error is already clear.]

Thus, for $n \geq 3$, we have that \begin{align} t^{n-2} & = 0 && \text{or} && t^2 - 14t + 33 && = 0\\ t & = 0 && \text{or} && (t-11)(t-3) && = 0\\ t & = 0 && \text{or} && t -11 && = 0 && \text{or} && t-3 && = 0\\ t & = 0 && \text{or} && t && = 11 && \text{or} && t && = 3\\ \end{align} and we see that the roots are $0$, $\mathbf{+}11$, and $\mathbf{+}3$. I presume that $t = 0$ ends up having no role (since $0^n$ is just $0$ and it is just extra hassle to add it when it does not help).

Charles Baker
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0

Here is a differential equations solution:

$f(x)=14f(x-1)-33f(x-2)$

With a bit of work we can find,

$f(x)=c_{1}(7-\sqrt82)^x+c_{2}(7+\sqrt82)^x$.

We just need to find the constants $c_{1}$ and $c_{2}$. Luckily, you've provided us with initial conditions I can use to find $c_1$ and $c_2$.

$a_0=-24$

$a_1-200$

Or,

$f(0)=-24$

$f(1)=-200$

To solve this equation we must solve,

$c_{1}+c_{2}=-24$

And,

$-200=(7-\sqrt82)c_{1}+(7+\sqrt82)c_{2}$

These equations as tedious to solve so I plugged them into my scientific calculator and apparently:

$c_{1}=-2(3\sqrt82-4)\sqrt\frac{2}{\sqrt41}$

And,

$c_{2}=-12-8\sqrt\frac{2}{\sqrt41}$

If you plug the constants into the sequence you'll get your answer. By the way, $f(x)=a_{n}$.

Aleksandar
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