已知数列{An}的前n项和公式Sn=32n-n^2,求新数列{/An/}的n项和Tn

an=sn-sn-1
=32n-n^2-32n+32+n^2-2n+1
=-2n+33
-2n+33>0 n<33/2≈16
即|an|前十六项和为
31+29+27+25+……+1=256
当n>16时，Tn =(1+2n-33)*(n-16)/2=n^2-32n+264
当n<=16时，Tn =(31+33-2n)*n/2=-n^2+32n

An=Sn-S（n-1）
=32n-n^2-32n+32+n^2-2n+1
=-2n+33
-2n+33>0 n<33/2，n≤16 。
所以
n≤16时，
|An|=|-2n+33 |=-2n+33
Tn =|32n-n^2|=-n^2+32n
当n＞16时，
|An|=|-2n+33 |=2n-33
Tn =|Sn-S16|+|S16|
=|32n-n^2-32*16+16^2|+32*16-16^2
=|32n-n^2-16^2|+16^2
=|-(n-16)^2|+16^2
=(n-16)^2+16^2
=n^2-32n+512

当n≤16时，Tn =-n^2+32n
当n＞16时，Tn =n^2-32n+512 。