Question

已知数列an,an=4n-32,求an的绝对值的前n项和

Answer 1

当1<=n<=7时，an<0；当n>=8时，an>=0
所以当1<=n<=7时
Sn=|a1|+|a2|+...+|an|

=-(a1+a2+...+an)
=-(4-32+4n-32)*n/2
=30n-2n^2
当n>=8时
Sn=|a1|+|a2|+...+|a7|+|a8|+...+|an|
=-(a1+a2+...+a7)+(a8+...+an)
=(a1+a2+...+an)-2*(a1+a2+...+a7)
=(4-32+4n-32)*n/2-2*(4-32+28-32)*7/2
=2n^2-30n+224

Answer 2

4n-32>0
n>8
得出
a1,a2,...,a7<0, a8=0, a8以后都是 大于 0
case 1: n≤8
|an| = -an = 32-4n
Sn
=|a1|+|a2|+...+|an|
=(1/2)n(28+32-4n)
= n(30-2n)
case 2: n>8
Sn
=|a1|+|a2|+...+|an|
=-(a1+a2+...+a8)+(a9+a10+a11+...+an)
=112 + (1/2)(n-8)(4+4n-32)
=112 + (1/2)(n-8)(4n-28)
=112 + 2(n-8)(n-7)
=2n^2-30n +224