Question

已知数列{an}的前N项和sn=(9n-n^2)/4

Answer 1

sn=(9n-n^2)/4

sn-1=(9n-9-n^2+2n-1)/4=(11n-10-n^2)/4

an=(-2n+10)/4=(5-n)/2

n5 tn=5+(n-5)/2+(n-5)(n-6)/4=(n^2-9n+40)/4

n^2-9n+40=(n-9/2)^2+79/4

n=1时 9n-n^2=8为最小值

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