Question:
Solve the following recurrance relation:
$\ a_n = 3a_{n-1} $
$\ a_1 = 4 $
But for the life of me, I can't understand why the answer is
$\ 3^{n-1}a_1 $
My workings so far...
$\ a_n = 3a_{n-1} $
$\ = 9a_{n-2} = 3^2a_{n-2} $
$\ = 3^2(3a{n-3}) = 3^3a_{n-3} $
$\ = 3^3(3a_{n-4}) = 3^4a_{n-4} $
$\ \vdots $
$\ \therefore a_n = 3^{???}a_1 $
I just simply do not see how you can flip $\ n-1 $ and the exponent 1 and get $\ 3^{n-1}a_1 $, could someone please help to explain why?
Thanks in advance!
We look for a solution in form of $a_n =\lambda ^n$
Plugging in the relation we get $$\lambda^ n =3\lambda^{n-1}$$ which results in $\lambda=3$
Thus with $a_n =c3^ n$ and $a_1=4$ we get $$a_n =4(3^{n-1})$$