Your first question: sn+1-3n+ 1 = 2sn+3n-3n+1(that is, both sides are reduced by 3n+1).

Sn+1-3n+1= 2sn+3n (1-3) (i.e. extracting 3n).

sn+ 1-3^n+ 1=2sn+3^n*(-2)

sn+ 1-3^n+ 1=2(sn-3^n)

The second question: I am too tired to play. Wait a minute. What is the answer? I haven't thought of that yet. I'll do the math again. In fact, the answer is not necessarily good. Maybe there are other ways to do it.

Stupid. I just understand now. Here's the thing:

bn=sn-3^n bn+ 1=sn+ 1-3^(n+ 1)

And because sn+1-3 n+1= 2 (sn-3 n)

So bn/bn+ 1=2.

b 1 = s 1-3 = a 1-3 = a-3

Can prove

Bn is the geometric series BN = b1q (n-1) = (a-3) 2n-1.

You got it?