设an=根号n+根号(n+1),求Sn=1/a1+1/a2+1/a3+...+1/an

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设an=根号n+根号(n+1),求Sn=1/a1+1/a2+1/a3+...+1/an
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设an=根号n+根号(n+1),求Sn=1/a1+1/a2+1/a3+...+1/an
设an=根号n+根号(n+1),求Sn=1/a1+1/a2+1/a3+...+1/an

设an=根号n+根号(n+1),求Sn=1/a1+1/a2+1/a3+...+1/an
an=根号n+根号(n+1),
1/an=1/(根号n+根号(n+1))
=[根号(n+1)-根号n]/[(根号n+根号(n+1)(根号(n+1)-根号n)]
=[根号(n+1)-根号n]/(n+1-n)
=根号(n+1)-根号n
1/a1=根号2-1
1/a2=根号3-根号2
.
1/an=根号(n+1)-根号n
相加:1/a1+1/a2+1/a3+...+1/an=根号(n+1)-1

1/an=根号(n+1)-根号n
Sn=根号(n+1)-1

1/an分母有理化=√(n+1)-√n
所以Sn=√2-1+√3-√2+……+√(n+1)-√n
=√(n+1)-1

Sn=1/(1+√2)+1/(√2+√3)+1/(√3+√4)...+1/(√n+√n+1)
分母有理化得:=√2-1+√3-√2+√4-√3.......√n+1-√n=√n+1-1

如图

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