Let an+k =1/2 (a (n-1)+k)

The contrast solution is k=- 1/2.

So an-1/2 =1/2 (a (n-1)-1/2).

{an- 1/2} is a geometric series with 1/2 as the first term and 1/2 as the common ratio.

an- 1/2= 1-( 1/2)^n

So an = 3/2-( 1/2) n

1、

. a(n+ 1)=2an+n^2+2n+ 1==>； a(n+ 1)+(n+3)^2+2=2[an+(n+2)^2+2]

So an+(n+2) 2+2 is a geometric series with a common ratio of 2.

n & gtan+(n+2)2+2 = 2(n- 1)*(a 1+3 2+2)= 2(n- 1)* 13 at 1。

an= 13*2^(n- 1)-(n+2)^2-2

When n= 1, a 1= 13-9-2=2 meets the question.

So the general formula of the sequence {an} is an =13 * 2 (n-1)-(n+2) 2-2.

I hope I can help you, wish you progress in your study, and don't forget to adopt it!