Question

已知数列｛an｝的前n项和sn=32n-n^2 求Tn=|a1|+|a2|+...|an|求sn

Answer 1

sn=32n-n^2=33n-2*(n+1)n/2

=[33-2]+[33-4]+...+[33-2n]

∴an=33-2n

Tn=|a1|+|a2|+...|an|

=(|33-2|+|33-4|+...+|33-32|)+ (|33-34|+|33-36|+...+|2n-33|)

=1+3+...+31+...+(1+3+5+...+[2n-33])

=(1+31)*16/2+(1+2n-33)*(n-16)/2

=256+(n-16)(n-8)