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
问题描述:
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
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