数列{An}的前n项和为Sn,若An=1/n(n+1),则S5=?

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数列{An}的前n项和为Sn,若An=1/n(n+1),则S5=?
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数列{An}的前n项和为Sn,若An=1/n(n+1),则S5=?
数列{An}的前n项和为Sn,若An=1/n(n+1),则S5=?

数列{An}的前n项和为Sn,若An=1/n(n+1),则S5=?
不要着急 要慢慢观察 这个是有规律可循的:拆项
an=1/n(n+1)=1/n-1/(n+1)
a1=1/2 a2=1/2-1/3
a3=1/3-1/4 a4=1/4-1/5
a5=1/5-1/6
S5=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6=5/6

裂项法计算
an=1/n(n+1)=1/n-1/(n+1)
a1=1/2
a2=1/2-1/3
a3=1/3-1/4
a4=1/4-1/5
a5=1/5-1/6
S5=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6
=5/6

an=1/n(n+1)
a1=1/2 a2=1/6 a3=1/12 a4=1/20 a5=1/30
S5=1/2+1/6+1/12+1/20+1/30
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6
=5/6