A function f is defined in such a way that f(1)=2, and for each positive integer n1, f(1)+f(2)+f(3)+...+f(n)=n^2f(n), evaluate the exact value of f(1000).

Question

A function f is defined in such a way that f(1)=2, and for each positive integer n>1, f(1)+f(2)+f(3)+...+f(n)=n^2f(n), evaluate the exact value of f(1000).

kinggeorge · Answer

I haven't proved anything, but I have noticed a striking pattern. For values above $n=2$, it appears that $$f(n)={f(1)\over T_n}$$where $T_n$ is the $n$-th triangular number. This would give $$\large f(1000)={2 \over {1000^2-1000 \over 2}}={1\over 999000}$$

asnaseer · Answer

You can rearrange to get this relationship:$$\begin{align}
f(1)+f(2)+...+f(n)&=n^2f(n)\
f(1)+f(2)+...+f(n-1)&=n^2f(n)-f(n)=(n^2-1)f(n)\
(n-1)^2f(n-1)&=(n^2-1)f(n)\
(n-1)f(n-1)&=(n+1)f(n)\
f(n)&=\frac{n-1}{n+1}f(n-1)
\end{align}$$

asnaseer · Answer

which then leads to....

asnaseer · Answer

$$\begin{align}
f(n)&=\frac{n-1}{n+1}f(n-1)\
&=\frac{n-1}{n+1}\frac{n-2}{n}f(n-2)\
&=\frac{n-1}{n+1}\frac{n-2}{n}\frac{n-3}{n-1}f(n-3)\
&...\
&=\frac{n-1}{n+1}\frac{n-2}{n}\frac{n-3}{n-1}...\frac{1}{3}f(1)\
&=\frac{2(n-1)!}{(n+1)!}f(1)\
&=\frac{2}{n(n+1)}f(1)
\end{align}$$

asnaseer · Answer

from which you should be able to calculate f(1000)

kinggeorge · Answer

So the pattern I noticed was correct!

asnaseer · Answer

but I get:$$f(1000)=\frac{2}{1000*1001}*2=\frac{1}{250250}$$

asnaseer · Answer

I believe your pattern results in:$$f(1000)={2 \over {1000^2-1000 \over 2}}={1\over 249750}$$

kinggeorge · Answer

I just didn't use the formula for triangular numbers >.< It should be $$\large {2 \over {1000^2+1000 \over 2}}$$And I calculated it wrong.

asnaseer · Answer

ah yes! - you're right - I should have spotted that as well :)