Question

已知sn=32n-n^2求等差数列|an|的前n项和sn

Answer 1

sn=32n-n^2;

n=1,sn=31;a1=31

d=-2

ai=31-2*i;

前n项和公式为：Sn=na1+n(n-1)d/2 =31n+n(n-1)d/2 =32n-n^2;

n(n-1)d/2 =n-n^2

d=-2,

ai=31-2*i;