Sn=n(n+2) 求1/S1+1/S2+...Sn=
来源:百度知道 编辑:UC知道 时间:2024/06/08 06:06:00
Sn=n(n+2) 1/Sn=1/[n(n+2)]=(1/2)[1/n-1/(n+2)]
1/S1+1/S2+...Sn=(1/2)[1-1/3+1/2-1/4+1/3-1/5+…+1/(n-1)-1/(n+1)+1/n-1/(n+2)]
=(1/2)[1+1/2-1/(n+1)-1/(n+2)]
=3/4-(2n+3)/[2(n+1)*(n+2)]
Sn=n(n+2)
1/S1=1/(1*3)=0.5(1-1/3)
所以原式等于0.5(1-1/3+1/2-1/4+1/3-1/5....+1/n-1/n+2)=0.5(1+1/2-1/n+1-
1/n+2)=3n2+13n+12/{2(n+1)(n+2)}=3/4-(2n+3)/2(n+1)(n+2)
Sn=1/2(1/n-1/n+2)
原式=1/2(1-1/3+1/2-1/4+1/3-1/5+…+1/(n-1)-1/(n+1)+1/n-1/(n+2))
=1/2(1+1/2-1/(n+1)-1/(n+2))
=3/4-(2n+3)/(n+1)(n+2)
3/4-(2n+3)/2(n+1)(n+2)
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