1, a 1=2, a2=3, a3=4, a4=5, a5=6, which proves that an is the arithmetic progression of arithmetic 1, that is, an=n+ 1.

When n=k, AK+1= AK 2-kak+1holds, that is, AK+1= k+1= k+2 = (k+/kloc-0

Then when n=k+ 1, a (k+2) = k+2+1= k+3; A (k+ 1) 2 = (k+2) 2, then a (k+1) 2-(k+1) a (k+1)+1= (k+)

2. When the first subproblem a 1=3, an=n+2 assumes that n = k and AK ≥ k+2 holds.

The proof method is the same as above, the basic principle of mathematical induction.

The second question draws a conclusion from the first question.