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- Feb 14, 2012

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- Thread starter anemone
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- Thread starter
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- #1

- Feb 14, 2012

- 3,964

- Jan 25, 2013

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let f(1)=k=1996

\(\displaystyle f(1)+f(2)+\cdots+f(n-1)=n^2f(n)-f(n)=(n-1)^2f(n-1)\)

$\therefore f(n)=\dfrac{n-1}{n+1}\times f(n-1)$

$ f(1996)=\dfrac {(k-1)(k-2)(k-3)-------(3)(2)(1)}{{(k+1)}(k)(k-1)----(5)(4)(3)}\times f(1)=\dfrac {2}{k+1}$

$=\dfrac {2}{1997}$