数列{An}满足A1=1/2,A1+A2+...+An=n^2*An,用数学归纳法证明An=1/[n(n+1)]

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数列{An}满足A1=1/2,A1+A2+...+An=n^2*An,用数学归纳法证明An=1/[n(n+1)]
数列{An}满足A1=1/2,A1+A2+...+An=n^2*An,用数学归纳法证明An=1/[n(n+1)]

数列{An}满足A1=1/2,A1+A2+...+An=n^2*An,用数学归纳法证明An=1/[n(n+1)]
当n=1时,A1=1/2
An=1/(1×2)=1/2
原式成立
设当n=k时,Ak=1/(k(k+1))
则n=k+1时
A1+A2+……+Ak+A(k+1)=(k+1)^2×A(k+1)
又∵A1+A2+……+Ak=k^2×Ak=k^2×1/(k(k+1))=k/(k+1)
两式相减
A(k+1)=(k+1)^2×A(k+1)-k/(k+1)
(k^2+2k)A(k+1)=k/(k+1)
A(k+1)=1/((k+1)(k+2))
原式也成立.
An=1/(n(n+1))

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