Strong inductive proof for recursive sequence given Explicit formula I am stuck on this
Show that the recursively defined sequence $a_n=2·a_{n-1}+3·a_{n-2}, a_1=-1, a_2=13$ has the explicit formula $〖a_n=3^n+4(-1)^n〗$
I can prove the base cases no problem but I keep getting stuck on the transformation
so I have $a_{k+1}=2·a_{k} + 3·a_{k-1}$
=$2(3^k+4(-1)^k)+3( 3^{k+1})+4(-1)^{k-1})$
=$2*3^k+8*(-1)^k+3*3^{k-1}+12*(-1)^{k-1}$
=$$2*3^k+8*(-1)^k+3^k- 12(-1)^k$$
=$$3*3^k-4(-1)^k$$
How do I get that -4(-1)^k to turn into +$4(-1)^{k+1}$?
Did I goof the alegbra somewhere?

You want to show:
$$3^{n}+4(-1)^{n} = 2[3^{n-1}+4(-1)^{n-1}] + 3[3^{n-2}+4(-1)^{n-2}]$$
Simplifying from the left we have:
$$2[3^{n-1}+4(-1)^{n-1}] + 3[3^{n-2}+4(-1)^{n-2}] = 2[3^{n-1}] + 8[(-1)^{n-1}] + [3^{n-1}] + 12[(-1)^{n-2}] =$$
$$=3[3^{n-1}] - 8[(-1)^{n-2}] + 12[(-1)^{n-2}] = 3^{n} + 4(-1)^{n-2} = 3^{n} + 4(-1)^{n-2}$$
