设数列1/(1+根2),1/(根2+根3),……,1/(根n+根(n+1))的前n项和为Sn,求Sn
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设数列1/(1+根2),1/(根2+根3),……,1/(根n+根(n+1))的前n项和为Sn,求Sn
![设数列1/(1+根2),1/(根2+根3),……,1/(根n+根(n+1))的前n项和为Sn,求Sn](/uploads/image/z/6661358-62-8.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%971%2F%EF%BC%881%2B%E6%A0%B92%EF%BC%89%2C1%2F%EF%BC%88%E6%A0%B92%2B%E6%A0%B93%EF%BC%89%2C%E2%80%A6%E2%80%A6%2C1%2F%EF%BC%88%E6%A0%B9n%2B%E6%A0%B9%EF%BC%88n%2B1%EF%BC%89%EF%BC%89%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E6%B1%82Sn)
an=1/(√n+√n+1)=√(n+1)-√n
故Sn=a1+a2+……+an
=√2-√1+√3-√2+……+√(n+1)-√n
=√(n+1)-1
故Sn=a1+a2+……+an
=√2-√1+√3-√2+……+√(n+1)-√n
=√(n+1)-1
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