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

[anemone]

If $$f(1)=1996$$, $$f(1)+f(2)+\cdots+f(n)=n^2f(n)$$, find $$f(1996)$$.

 

[Albert1]

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}$
$=\dfrac {2}{1997}$