Question

求一道高二数列题,要解法~~

设等差数学{a(n)}前n项和为Sn,已知a(3)=12,S(12) > S(13),问S(1),S(2),S(3),...,S(n)中,哪一个最大?试说明理由 括号内为下标.

Answer 1

S12>S13
a3=12
a1+2d=12 => a1=12-2d
S12=6*(a1+a12)=6*(2a1+11d)=6*(24-4d+11d)=144+7d
S13=S12+a13=144+7d+12-2d+12d=156+17d
144+7d>156+17d
=> 10d<-12
d<-6/5
Sn=a1+[n(n-1)d/2]=12-2d+[n(n-1)d/2]=[(n^2-n+4)d/2]+12=[d/2*[(n-1/2)^2+7/4]=12
n>1/2递减
所以S1最大。

Answer 2

S(12) > S(13),说明公差d<0
同时可得，a1,a2,....a12>0,a13,a14,a15,.....<0.
S12最大！

Answer 3

258299054

Sn=a1+[n(n-1)d/2]=12-2d+[n(n-1)d/2]=[(n^2-n+4)d/2]+12=[d/2*[(n-1/2)^2+7/4]=12

错了！
Sn= n*a1+[n(n-1)d/2]=........