Question

已知数列{an}的前n项和为Sn=n^2-8n,（1）求数列{|an|}的通项公式（2）求数列{|an|}的前n项和Tn

Answer 1

1.an = Sn - Sn-1 = n^2 -8n -(n-1)^2 + 8(n-1)

= n^2 -8n - n^2 + 2n -1 + 8n -8

= 2n-9

|an| = |2n -9|

= 9 - 2n (n < 5)

2n -9 (n > 4)

2.要根据n的值讨论,现在有事,呆会有空再做.

n < 5时,Tn = 9n -2(1+2+...+n)

= 9n -n(n+1)

T4= 9*4 -4*5 = 16

n > 4 时,Tn = T4 + |a5| + |a6| + ...+ |an|

= 16 + 2(5 + 6 + ...+ n) -9(n-4)

= 16 + (5 + n)(n-4) -9n +36

= n^2 + n -20 -9n +52

= n^2 -8n +32