MHB Find f(1996): f(1)=1996, f(1)+f(2)+\cdots+f(n)=n^2f(n)

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The function f is defined such that f(1) equals 1996, and the relationship f(1) + f(2) + ... + f(n) = n^2f(n) holds. By manipulating the equation, it is derived that f(n) can be expressed as f(n) = (n-1)/(n+1) * f(n-1). This recursive relationship leads to the calculation of f(1996) as a product involving decreasing integers and the initial value f(1). Ultimately, f(1996) is determined to be 2/1997.
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If $$f(1)=1996$$, $$f(1)+f(2)+\cdots+f(n)=n^2f(n)$$, find $$f(1996)$$.
 
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anemone said:
If $$f(1)=1996$$, $$f(1)+f(2)+\cdots+f(n)=n^2f(n)$$, find $$f(1996)$$.
let f(1)=k=1996
$$f(1)+f(2)+\cdots+f(n-1)=n^2f(n)-f(n)=(n-1)^2f(n-1)$$
$\therefore f(n)=\dfrac{n-1}{n+1}\times f(n-1)$
$ 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}$
$=\dfrac {2}{1997}$
 
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