已知数列{an}的前n项和Sn=n^2-2n+3,求数列{an}的通项an,并判断数列{an}是否为等差数列.

由已知：S n =n 2 -2n+3,所以,S n-1 =(n-1) 2 -2(n-1)+3=n 2 -4n+6,两式相减,得：a n =2n-3(n 2),而当n=1时,a 1 =S 1 =2,所以a n = 2（n=1）an=2n-3（n≥2）.又a 2 -a 1≠ a 3 -a 2 ,故数列{a n }不是等差数列....