0
$\begingroup$

solve recurrence relation $a_n = a_{n–1} + 12 a_{n–2}$, where $a_0 = 1$ and $a_1 = 11$ and Verify, using Principle of Mathematical Induction, that $a_n = (-1)(-3)^n + (2)(4^n)$.

ans: i have done so far.. put $a_n=b_n$
$a_n-a_{n-1}-12a_{n-2}=0$
$b_n-b_{n-2}-12a_{n-2}=0$
$b^2-b-12=0$, $b=-3,4$

$\endgroup$
  • $\begingroup$ Why are you putting $a_n = b_n$? What purpose does having two notation for the same terms mean? $\endgroup$ – fleablood Oct 22 '15 at 16:44
  • $\begingroup$ A "classic" induction proof would show it's true for initial values of n (so n = 2 here), then assuming it's true for some n, show it's also true for n + 1. I haven't followed through with the proof so it may end up not working out, but that's how I'd start here. $\endgroup$ – blm Oct 22 '15 at 16:45
  • 1
    $\begingroup$ So then the general solution is $a_n = k_1(-3)^n + k_2(4)^n$, and use initial conditions $a_0 = 1, a_1 = 11$ to get the general $a_n = (-1)(-3)^n + (2)(4^n)$. I am assuming that you can take care of the induction $\endgroup$ – Shailesh Oct 22 '15 at 16:45
  • $\begingroup$ I don't know how to politely say this but everything about this is completely wrong. $a_n \ne a^n$. This isn't trying to solve an equation. $a_n$ simply means that it is the nth term in a sequence of numbers. The first, $a_0 = 1$ and the next, $a_1$ is 11, so the third is $a_2 = a_1 + 12a_0 = 11 + 12 = 23$. And so on to get a sequence of numbers {1,11,23,a_3, a_4, .......} You need to do a "proof by induction" to show that the $a_n$ term of these numbers is actually equal to $(-1)(-3)^n + (2)(4^n)$ $\endgroup$ – fleablood Oct 22 '15 at 16:51
  • $\begingroup$ Maybe you meant $a_n = b^n$ ? $\endgroup$ – Joel Cohen Oct 22 '15 at 16:52
0
$\begingroup$

$a_{n+1} = a_n + 12a_{n-1}=$

$[(-1)(-3)^n + (2)(4^n)] + 12[(-1)(-3)^{n-1} + (2)(4^{n-1})]=$

.....

$(-1)(-3)^{n+1} + (2)(4^{n+1})$

Fill in the ...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy