设正数a1,a2,a3,···an成等差数列,求证:1/(根号a1+根号a2)+1/(根号a2+根号a3)+···+1/(根号an+a(n-1)==n/(根号a1+a(n+1)

设正数a1,a2,a3,···an成等差数列,求证:1/(根号a1+根号a2)+1/(根号a2+根号a3)+···+1/(根号an+a(n-1)==n/(根号a1+a(n+1)
设正数a1,a2,a3,···an成等差数列,求证:1/(根号a1+根号a2)+1/(根号a2+根号a3)+···+1/(根号an+a(n-1)
==n/(根号a1+a(n+1)

设正数a1,a2,a3,···an成等差数列,求证:1/(根号a1+根号a2)+1/(根号a2+根号a3)+···+1/(根号an+a(n-1)==n/(根号a1+a(n+1)
证明,假设等差数列的公差为d.
因为
1/(根号a1 + 根号a2)
= (根号a2 - 根号a1) / (a2-a1)
= (根号a2 - 根号a1) / d
同理可得
1/(根号a2 + 根号a3)
= (根号a3 - 根号a2) / d
所以类似的有
1/(根号a1+根号a2)+1/(根号a2+根号a3)+···+1/(根号a(n+1)+根号an)
=( (根号a2 - 根号a1) + (根号a3 - 根号a2) + ...+ (根号a(n+1) - 根号an )/d
= (根号a(n+1) - 根号a1)/d
= n(根号a(n+1) - 根号a1) / (nd)
= n(根号a(n+1)-根号a1) / (a(n+1) - a1)
= n / (根号a(n+1) + 根号a1)

设公差为d
1/(根号a1+)+1/(根号a2+根号a3)+···+1/(根号an+a(n-1)
=(√a2-√a1)/(a2-a1)+(√a3-√a2)/(a3-a2)+....+[√an-√a(n-1)]/[(an-a(n-1)]
=(1/d){(√a2-√a1)+(√a3-√a2)+....+[√an-√a(n-1)]}
=(1/d)(√an-√a1)

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设公差为d
1/(根号a1+)+1/(根号a2+根号a3)+···+1/(根号an+a(n-1)
=(√a2-√a1)/(a2-a1)+(√a3-√a2)/(a3-a2)+....+[√an-√a(n-1)]/[(an-a(n-1)]
=(1/d){(√a2-√a1)+(√a3-√a2)+....+[√an-√a(n-1)]}
=(1/d)(√an-√a1)
=(1/d)(an-a1)/(√an+√a1)
=(n-1)/(√an+√a1)
与你的不合啊！

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an=a1+(n-1)d
2楼正解
按照题目规律来看 左式总共有n-1 项
可能此题为盗版书上的题吧 少了一项 +1/（根号an+根号a(n+1)）

设该等差数列公差为d
当d=0时，该等差数列公差为常数列，a1=a2=a3=···=an=a(n+1)
1/(根号a1+根号a2)+1/(根号a2+根号a3)+···+1/(根号an+a(n+1))=n/(根号an+a(n+1))=n/(根号a1+a(n+1))
当d不等于0时
1/(根号a1+根号a2)+1/(根号a2+根号a3)+···+1/(根号an+a(n+...

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设该等差数列公差为d
当d=0时，该等差数列公差为常数列，a1=a2=a3=···=an=a(n+1)
1/(根号a1+根号a2)+1/(根号a2+根号a3)+···+1/(根号an+a(n+1))=n/(根号an+a(n+1))=n/(根号a1+a(n+1))
当d不等于0时
1/(根号a1+根号a2)+1/(根号a2+根号a3)+···+1/(根号an+a(n+1))
=(根号a1-根号a2)/[(根号a1+根号a2)根号a1-根号a2)]+(根号a2-根号a3)/[(根号a2+根号a3)(根号a2-根号a3)]+···+[根号an根号-a(n+1)]/[(根号an+根号a(n+1))(根号an-根号a(n+1)]
=(根号a1-根号a2)/(a1-a2)+(根号a2-根号a3)/(a2-a3)+···+[根号an-根号a(n+1)]/(an-a(n+1))
=(根号a1-根号a2)/(-d)+(根号a2-根号a3)/(-d)+···+[根号an--根号a(n+1)]/(-d)
=[(根号a1-根号a2)+(根号a2-根号a3)+···+根号an-根号a(n+1)]/(-d)
=(根号a1-根号a(n+1))/(-d)=(根号a1-根号a(n+1))(根号a1+根号a(n+1))/[(-d))(根号a1+根号a(n+1))]
=(a1-a(n+1))/[(-d))(根号a1+根号a(n+1))]=-nd/[(-d))(根号a1+根号a(n+1))]=n/(根号a1+根号a(n+1))
综上等式成立

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