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数学
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数列{an}的前n项和Sn=32n-n^2,求{|an|}的前n项和Pn
人气:415 ℃ 时间:2020-06-05 14:13:35
解答
an=sn-sn-1
=32n-n^2-32n+32+n^2-2n+1
=-2n+33
-2n+33>0 n<33/2≈16 即|an|前十六项和为
31+29+27+25+……+1=256
当n>16时,Pn=(1+2n-33)*(n-16)/2=n^2-32n+264
当n<=16时,Pn=(31+33-2n)*n/2=-n^2+32n
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