设数列的前几项和Sn=2n^2-4n+1(n属于N+),求数列的通项公式
人气：143 ℃ 时间：2020-06-15 21:49:15
解答
a(1) = S(1) = 2 - 4 + 1 = -1S(n-1) = 2(n-1)^2 - 4(n-1) + 1 = 2n^2 - 4n + 2 - 4n + 4 + 1 = 2n^2 - 8n + 7.a(n) = S(n) - S(n-1) = 2n^2 - 4n + 1 - [2n^2 - 8n + 7]= 4n - 6.a(1) = -1,a(n) = 4n - 6,n = 2,3,....
推荐
猜你喜欢
