Find an explicit formula for the recurrence relation $a_n = 5_{an−1} + 6_{an−2}$ with initial conditions $a_1$ = 5 and $a_2$ = 11.

Question

So here is what I did

$a_n = 5_{an−1} + 6_{an−2}$

$a_n - 5_{an−1} - 6_{an−2} = 0$

which becomes

$t^2 -5t -6$

$(t+1)(t-6)$

$t = -1$ and $t= 6$

then we have

$a_n = $$-1^nB + 6^nD$

and so using

$a_1 = 5 \ and \ a_2 = 11$

$5 = $$ -1B + 6D$

$11= $$ 1B + 36D$

$16 = $$ 42D$

so D = $ \frac 8{21}$

and following the same process

B = $ \frac {35}{16}$

$a_n = $$-1^n $$ \frac {35}{16}$ + $6^n$$ \frac 8{21}$

Is this correct? It worked out so not nicely number wise I wasn't sure.

Answer 1

Your overall process is fine, though a bit of the work is wrong. My nitpicks and some words of advice:

• At one point, you have $a_n = -1^n + ...$. That should be $(-1)^n$; $-1^n = -1$, the parentheses are important.

• You made an arithmetic error in finding $B$. Plugging in $D$ to the equation for $a_0$, you would get $5 = -B + 6(8/21)$, which would yield a solution of $-19/7$, not $35/16$.

• A good paranoia check for these sorts of things: ensure that your explicit formula gives you the initial conditions back (so if you plug in $n=1$, you should get $a_1 = 5$ back). You can also use the recurrence to get some later non-initial values by hand and verify those; if they're not matching your explicit formula for $a_n$, something's wrong.

• Also, don't be shocked if the numbers don't "look nice." Even this is comparatively nice to some solutions I've seen; for example, the Fibonacci relation has the solution $$a_n = \frac{1}{\sqrt5} \left( \frac{1 + \sqrt 5}{2} \right)^n - \frac{1}{\sqrt5} \left( \frac{1 - \sqrt 5}{2} \right)^n$$ and the relation itself ($a_n = a_{n-1} + a_{n-2}$) is really simple-looking!