解:设等差数列{a n }的公差为d,前n项的和为S n ,所以S n = na 1 + n(n – 1)d/2 = (d/2)n 2 + (a 1 – d/2)n = an 2 + bn,所以 b = a 1 – d/2 (b是等差数列{a n }的首项a 1 与公差的一半 之差); 举例:对于等差数列{a n }:a n = 2n + 1,首项a 1 = 3,公差d = 2,此时等差数列{a n }前n项的和S n = n*3 + n(n – 1)d/2 = 3n + n 2 – n = n 2 + 2n,b = 2 = a 1 – d/2 。