Discrete math and recursion problem.

Question

I was recently reading up examples on recursion and how it relates to induction and there's this question I am not sure about.

Q: Let $$b_1=3$$ $$b_n=n(n+2)$$

From that question I wanted to do the n+1th step as well for the recursion process and i got this:

$$b_{n+1} = b_n+3$$

however, that is wrong and apparently it is missing a $+2n$ in it:

$$b_{n+1}=b_n+2n+3$$

Can someone please explain how the book got $2n$?

Answer 1

$$b_n=n(n+2)$$ $$b_{n+1}=(n+1)(n+3)=n(n+3)+1(n+3)=n(n+2)+n+n+3=b_n+2n+3$$

Answer 2

$b_{n+1}=(n+1)(n+3)=n^2+4n+3=n(n+2)+2n+3=b_n+2n+3$