I am working through some of the problems in the appendix of my algorithms book...

They are showing that $\displaystyle \sum_{k=0}^{n}3^k$ is $\displaystyle O(3^n)$ or $\displaystyle \sum_{k=0}^{\n}3^k <= c3^n$ for some constant c

The initial condition was shown to be true as long as c >= 1. They assume that the bound holds for n and then show that it holds for n+1

$\displaystyle \sum_{k=0}^{n+1}3^k = \sum_{k=0}^{n}3^k +3^{n+1} <= c3^n + 3^{n+1} = (\frac{1}{3} + \frac{1}{c})c3^{n+1} <= c3^{n+1} $

as long as $\displaystyle (\frac{1}{3} + \frac{1}{c}) <= 1 $ or $\displaystyle c>=3/2$

What I don't get is $\displaystyle (\frac{1}{3} + \frac{1}{c})c3^{n+1}$ I see that I can distribute the $\displaystyle c3^{n+1}$ to get the previous result. But Why do we want this form?

Is it because I want to express $\displaystyle c3^n + 3^{n+1}$in a way that I can make it comparable to $\displaystyle c3^{n+1}$?