设a1=1,a n+1=a n + 1/2,则数列{a n}的前n项之和为 A.（n^2+3n）/2 B.(n^2+n)/4 C.(n^2+n)/2
D..（n^2+3n）/4给出过程
人气：449 ℃ 时间：2020-09-05 13:29:23
解答
D..（n^2+3n）/4
a (n+1)=a n + 1/2
a (n+1)-a n =1/2
an是以1/2为公差的等差数列
an=a1+(n-1)d
=1+(n-1)*1/2
=n/2+1/2
sn=(a1+an)*n/2
=(1+n/2+1/2)n/2
=(n/2+3/2)*n/2
=(n^2+3n)/4
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