Question

已知等差数列{an}的前n项和是sn=32n一n*n，求{|an|}的前n项和sn

Answer 1

解：当n=1时
a1=s1=32*1-1*1=31
当n≥2时
an=Sn-S(n-1)
=32n-n²-[32(n-1)-(n-1)²]
=32n-n²-(32n-32-n²+2n-1)
=-2n+33
令an>0，则-2n+33>0
解得：n<33/2，即n最大为16
∴1°若n≤16
则|Sn|=(a1+an)n/2
=(31+33-2n)n/2
=32n-n^2
2°若n≥17
则|Sn|=|a1|+|a2|+……+|a16|+|a17|+|a18|+……+|an|
=16(|a1|+|a16|)/2+|a17|+|a18|+……+|an|
=8*(31+1)+1+3+……(2n-33)
=256+(1+2n-33)*(n-16)/2
=256+(n-16)(n-16)
=n²-32n+512

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Answer 2

是这样的吗:Sn=32n-n^2:
a1=S1=32-1=31
n>=2:an=Sn-S[n-1]=32n-n^2-32(n-1)+(n-1)^2=32-2n+1=33-2n
an<0,有n>33/2,即有当a16>0,a17<0
{|an|}的前n项和Sn
(1)n<=16,Sn=(a1+an)n/2=(31+33-2n)n/2=32n-n^2
n>17,Sn=(a1+...+a16)-(a17+a18+...+an)
=-(a1+...+a16+a17+...+an)+2(a1+...+a16)
=-(32n-n^2)+2(31+1)*16/2
=n^2-32n+256

Answer 3