Question

已知数列{an}的前n项和为Sn=n平方-3n,求（1） an

Answer 1

Sn = a1 + a2 + …… a(n-1) + a(n)

S(n-1) = a1 + a2 + …… + a(n-1)

所以

Sn - S(n-1) = an

an = n^2 - 3n - [(n-1)^2 - 3(n-1)]

= n^2 -3n - ( n^2 -5n + 4)

= 2n -4

(2)

an = 2n -4

a(n-1) = 2(n-1) - 4 = 2n -6

an - a(n-1) = 2n - 4 - (2n -6) = 2 为 常数

所以 an 是等差数列