数列 (8 21:16:6)数列{an}的前n项和为Sn,若an=1/n*(n+1) ,则S5等于多少?)
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数列 (8 21:16:6)数列{an}的前n项和为Sn,若an=1/n*(n+1) ,则S5等于多少?)
数列 (8 21:16:6)
数列{an}的前n项和为Sn,若an=1/n*(n+1) ,则S5等于多少?)
数列 (8 21:16:6)数列{an}的前n项和为Sn,若an=1/n*(n+1) ,则S5等于多少?)
an=[(n+1)-n]/n(n+1)
=(n+1)/n(n+1)-n/n(n+1)
=1/n-1/(n+1)
所以S5=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6
=1-1/6
=5/6
S5=1/1-1/2+1/2-1/3+1/3-1/4....+1/5-1/6=5/6
an=1/n(n+1)
=1/n-1/(n+1)
Sn=(1-1/2)+(1/2-1/3)+…+[1/(n-1)-1/n]+[1/n-1/(n+1)]
=1-1/(n+1)
=n/(n+1) 裂项相消法
S5=5/6
5/6