a2-a 1=2- 1= 1

[a (n+2)-a (n+1)]-[a (n+1)-an] = 2, which is a fixed value.

Sequence {a(n+ 1)-an} is a arithmetic progression, the first term is 1, and the tolerance is 2.

bn=a(n+ 1)-an

Sequence {bn} is an arithmetic series, the first term is 1, and the tolerance is 2.

(2)

a(n+ 1)-an = 1+2(n- 1)= 2n- 1

an-a(n- 1)= 2(n- 1)- 1

a(n- 1)-a(n-2)= 2(n-2)- 1

…………

a2-a 1= 1

total

an-a 1= 1+3+...+[2(n- 1)- 1]=(n- 1)？

an=a 1+(n- 1)？ = 1+(n- 1)？ =n？ -2n+2

When n= 1, a 1= 1? -2× 1+2= 1; When n=2, a2=2? -2×2+2=2, both of which satisfy the expression.

The general formula of the sequence {an} is an=n? -2n+2

There is nothing wrong with your problem-solving process, and the reason why you can't get the final result may be that you don't remember this formula:

1+3+...+(2n- 1)=n？