Hi.
Yes, you are total violating the forum rules.
Well, CB gave you a perfect answer, I'm going to quote him (you can read it here : http://www.mathhelpforum.com/math-he...tml#post327278 )
And you should give him "thanks" for his post, I already gave him reputation...
I quote again and explain the last obvious step
$\displaystyle
T(n) = 2T\left(\frac{n}{2}\right) + 2n - 1 $ (1)
so:
$\displaystyle
T(n/2) = 2T(n/4) + n - 1 $ (2)
you see what happened here? Just substitute n := n/2 in (1)
And now lets take a closer look to (1):
$\displaystyle
T(n) = 2T\left(\frac{n}{2}\right) + 2n - 1 $
Do you see the T(n/2)? You know T(n/2), it is '(2)'
Just use (2) in (1)
$\displaystyle
T(n) = 2*[2T(n/4) + n - 1 ]+ 2n - 1 $ ((2) in (1))
You get it now?
Yours
Rapha