p=1/(2+2^1/2)+1/(18^1/2+12^1/2)+.+1/[(n+1)n^1/2+n(n+1)^1/2](

mm转出事2022-10-04 11:39:542条回答

p=1/(2+2^1/2)+1/(18^1/2+12^1/2)+.+1/[(n+1)n^1/2+n(n+1)^1/2](n为正整数）证p小于1,大于0

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czc4896 共回答了19个问题 | 采纳率100%: 1/[(n+1)n^1/2+n(n+1)^1/2]
=[(n+1)√n-n√(n+1)]/n(n+1)
=1/√n-1/√(n+1)
p=1/(2+2^1/2)+1/(18^1/2+12^1/2)+.+1/[(n+1)n^1/2+n(n+1)^1/2]
=(1-1/√2)+(1/√2-1/√3)+...+(1/√n-1/√(n+1))
=1-1/√(n+1)
所以,0; 1年前

124629594 共回答了2个问题 | 采纳率: 1/[(n+1)n^1/2+n(n+1)^1/2]
=[(n+1)√n-n√(n+1)]/n(n+1)
=1/√n-1/√(n+1); 1年前

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