数学一题呀..等

Sn/n=9+(-1)(n-1)=10-n
Sn=n*(10-n)
所以An=Sn-S(n-1)
=n*(10-n)-(n-1)*(10-(n-1))
=10n-n^2-(n-1)*(11-n)
=10n-n^2+n^2-12n+11
=-2n+11

设{an}的通项公式为an=a1+(n-1)d
等差数列的N项和Sn=na1+1/2n(n-1)d
则Sn/n=a1+1/2(n-1)d=a1+(n-1)(d/2)
则Sn/n是以a1为首项，d/2为公差的等差数列
由题意可知
a1=9,d=-2
则an=9+(n-1)（-2）=11-2n

Sn/n=9-(n-1)=8-n
Sn=(8-n)n=8n-n^2
S(n-1)=-9+10n-n^2
an=Sn-S(n-1)=9-2n

an=9+(n-1)*(-1)=10-n