$5(2^{n−1} + 5 ·3^{n−1}) − 6(2^{n−2} + 5 · 3^{n−2}) = 2^{n−2}[10 − 6] + 3^{n−2}[75 − 30] = 2^{n−2} · 4 + 3^{n−2} · 9 · 5 = 2^n + 3^n · 5 $
There are enormous leaps in my understand between each section listed. The answer is there, but I don't understand how they arrived at it. Would anyone mind trying to clarify? Been out of College Algebra for quite a few years and I'm taking a upper graduate Discrete Math class that requires this proof for recurrence relations.
As the lowest power of $2$ is $n-2,$
we set $2^{n-2}=a\implies 2^{n-1}=2\cdot2^{n-2}=2a$
So, if we consider the terms containing the power of $2,$
$5(2^{n−1}) − 6(2^{n−2}) =5(2a)-6(a)=a(5\cdot2-6)=4a=2^2\cdot2^{n-2}=2^{(2+n-2)}=2^n$
Similarly, for the terms containing the power of $3$