题目

已知等差数列{An}的前n项和为Sn=2n²－10n,求(1)a1和a3的值；(2)求a5+a6+a7+a8；(3)求它的通项公式并判断它是否为等差数列.

提问时间：2020-11-04

答案

a(1) = s(1)= 2-10= -8.
s(n) = 2n^2 - 10n,
a(n+1) = s(n+1) - s(n) = [2(n+1)^2 - 10(n+1)] - [2n^2 - 10n] =2(2n+1) - 10 = 4n -8,
a(n) = 4(n-1) - 8 = 4n - 12.
{a(n) = 4(n-1)-8}是首项为a(1)=-8,公差为4的等差数列.
a(1) = -8,a(3) = 4*3 - 12 = 0.
a(5) + a(6) + a(7)+a(8) = 4[5+6+7+8] - 12*4 = 4[5+6+7+8-12] = 4*14 = 56