So we get a 2-3an = [a (n-1)] 2+3a (n-1).

So a 2-[a (n-1)] 2 = 3an+3a (n-1),

As can be known from the square difference formula

An 2-[a (n-1)] 2 = [an-a (n-1)] * [An+A(n- 1)]

that is

[An-A(n- 1)]*[An+A(n- 1)]= 3An+3A(n- 1)

Because all the items in the sequence {An} are positive numbers,

So An+A(n- 1) is not equal to 0,

So divide by An+A(n- 1) on both sides of the equation,

You get a -A(n- 1)=3.

I won't talk about the rest of the steps.