已知数列｛an｝的通项公式an=1+2+3+⋯n+/2,则｛an｝的前n项之和是多少（麻烦了,

Question

已知数列｛an｝的通项公式an=1+2+3+⋯n+/2,则｛an｝的前n项之和是多少（麻烦了,

Answer 1

∵ an＝1/[n*(n+2)]=1/2[1/n-1/(n+2)]
∴Sn＝a1+a2+a3+a4+a5+.+an-2+an-1+an
＝1/2[(1-1/3)+(1/2-1/4)+(1/3-1/5)+(1/4-1/6)+(1/5-1/7)+.+(1/(n-2)-1/n)+(1/(n-1)-1/(n+1))+(1/n-1/(n+2))]
＝1/2[1+1/2-1/(n+1)-1/(n+2)]
∴limSn=lim{1/2[1+1/2-1/(n+1)-1/(n+2)]}=1/2(1+1/2)=3/4
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