已f(1)+f(2)+...+f(n)=n²f(n), 若f(1)=1010,求f(2020)之值?

f(1) + f(2) + ... + f(n - 1) + f(n) = n² f(n)
f(1) + f(2) + ... + f(n - 1)           = (n - 1)² f(n - 1)

所以
(n - 1)² f(n - 1) + f(n) = n² f(n)
(n - 1)² f(n - 1) = n² f(n) - f(n)
(n - 1)² f(n - 1) = (n² - 1) f(n)
(n - 1)² f(n - 1) = (n + 1)(n - 1) f(n)
(n - 1) f(n - 1) = (n + 1) f(n)

可得
f(n)/f(n - 1) = (n - 1)/(n + 1)

考慮
f(2020)/f(2019) = 2019/2021
f(2019)/f(2018) = 2018/2020
f(2018)/f(2017) = 2017/2019
f(2017)/f(2016) = 2016/2018
...
f(4)/f(3) = 3/5
f(3)/f(2) = 2/4
f(2)/f(1) = 1/3

把以上等式相乘：
f(2020)/f(1) = (1)(2)/[(2020)(2021)]
f(2020)/(1０1０) = 1/[(1０1０)(2021)]
f(2020) = 1/2021