How is $2^{n-k-1} \cdot 3^k = (3/2)^k \cdot 2^{n-1}$? I'm currently reading through some litterature about difference equations, and i came across a relation i just can't seem to wrap my head around.
How come:
$2^{n-k-1}\cdot 3^k=(\frac{3}{2})^k\cdot 2^{n-1}$?
The original problem in the book is:
(1*) $y(n)=a^ny_0+\sum_{r=0}^{n-1}a^{n-r-1}g(r)$
Find a solution for the equation:
$x(n + 1) = 2x(n) + 3n$ , $x(1) = 0.5$
Solution From (1*), we have
(2) $x(n) = (\frac{1}{2})\cdot2^{n-1}+\sum_{r=1}^{n-1}2^{n-r-1}3^k$
(3) $x(n) = 2^{n-2}+2^{n-1}\sum_{r=1}^{n-1}(\frac{3}{2})^k$
(4) $x(n) = 2^{n-2}+2^{n-1}\frac{3}{2}(\frac{(\frac{3}{2})^{n-1}-1}{\frac{3}{2}-1})$
(5) $x(n) = 3^n -5\cdot2^{n-2}$
Although my question is more directed to the relation $2^{n-k-1}\cdot 3^k=(\frac{3}{2})^k\cdot 2^{n-1}$, a thorough explanation of the steps in the book would be appreciated as well. 
I'm truly sorry if this question has been asked before or is in any way irrelevant to the tags, but I'm pretty green to this topic, so the relevance of the tags might be off.

$$2^{n-k-1}\cdot 3^k=\frac{2^{n-1}}{2^k}\cdot 3^k=2^{n-1}\cdot \frac{3^k}{2^k} =\left(\frac{3}{2}\right)^k\cdot 2^{n-1}$$
