sn=32-n^2
sn-1=32-(n-1)^2
an=sn-sn-1
=32-n^2-[32-(n-1)^2]
=1-2n
1-2n1
n>1/2
a1=-1
sn=(-1+1-2n)*n/2
=-n^2
Tn=|sn|=|-n^2|=n^2
S12=354
前12項(xiàng)中偶數(shù)項(xiàng)的和與奇數(shù)項(xiàng)的和之比為27：32
偶數(shù)項(xiàng)和:354*32/(32+27)=192
奇數(shù)項(xiàng)和:354*27/(33+27)=162
a2=a1+d,a4=a3+d,a6=a5+d
所以a2+a4+...+a12=a1+a3+...+a11+6d
192-162=6d
6d=30
d=5
S12=(a1+a1+11d)*12/2
=(2a1+11*5)*6
=12a1+330
12a1+330=354
a1=2
an=a1+(n-1)d
=2+(n-1)5
=5n-3