SRL Archive - General

So I have
an = (2^n)-1 for n>=1
and I want to know is this recursive formula equivalent to it
an = 2*(an_-1) +1


When I commuted it they don't appear to be equal but the solutions manual says otherwise.
This is their solution: if an = (2^n) -1, then (an_-1) = (2^(n-1) -1, so 2*(an_-1) + 1 = 2(2^(n-1) - 1) + 1 = (2^n) -1

Explanation would be pleasant.

From: [XoL]
So I have
an = (2^n)-1 for n>=1
and I want to know is this recursive formula equivalent to it
an = 2*(an_-1) +1

Is the following true?
2^n-1=an?=2(an_-1)+1
2^n-1=an?=2(an_)-1
2^n=an?=2an_

I'm assuming 'an' is a single variable that is not the same as 'an_'. = is a guaranteed equals and ?= is a 'does it equal, if yes then your answer is yes'. Are my assumptions true?

Sorry was a bit unclear, an is the same variable as an_-1 (for the most part), the _ is subscript so I can right a recursive function (an_-1, meaning use the last answer)

Rewritten with subscript to read easier now :p

an_x = (2^n)-1 for n>=1 (GIVEN)
2*an_x-1 + 1 = an_x (DETERMINE)

an_x-1 = (2^(n-1))-1
(multiply 2 by both sides)
2*an_x-1 = 2*(2^(n-1))-2
(simplify)
2*an_x-1 = 2^n-2
(add 1 to both sides)
2*an_x-1+1 = 2^n-1
(substitute)
2*an_x-1+1 = an_x