The construction method of sequence, such as a(n+ 1)=3an+2, and the teacher said bn=an+k, is just a change.

Assuming that a(n+ 1)+k=t[a(n)+k] holds, the sequence b(n)=a(n)+k, b(n+ 1)=t*b(n) can be constructed. Obviously, b(n) is a geometric series. You can also get a (n) = b (n)-k.

Then solve a(n+ 1)+k=t[a(n)+k].

a(n+ 1)+k=t*a(n)+tk

a(n+ 1)= t * a(n)+tk-k = t * a(n)+(t- 1)k

So t = 3, (t- 1) k = 2, k = 2/(t- 1) = 1.

Therefore, we can construct b(n)=a(n)+ 1.