Given a function $f(x)$ such that: $f(1) + f(2) + f(3)+\cdots+f(n) = n^2f(n)$
Find the value of $f(2013)$. It is given that $f(1) = 2014$.
I tried attempting the question as a bottom-up DP, but soon realized the numbers are too irregular and large to deal with in a mechanical fashion.
On
Rule of thumb: if the current year shows up in a problem, it is probably a red herring.
Second rule of thumb: looking for a pattern will help you decide what to prove.
$$f(1) + f(2) = 4 f(2) \implies f(2) = \frac{f(1)}{3}$$ $$f(1) + \frac{f(1)}{3} + f(3) = 9 f(3) \implies f(3) = \frac{f(1)}{6}$$ $$f(1) + \frac{f(1)}{3} + \frac{f(1)}{6} + f(4) = 16 f(4) \implies f(4) = \frac{f(1)}{10}$$ $$f(1) + \frac{f(1)}{3} + \frac{f(1)}{6} + \frac{f(1)}{10} + f(5) = 25 f(5) \implies f(5) = \frac{f(1)}{15}$$
From here you should be able to make an educated guess about the value of $f(2013)$ and a formal proof should not be hard to write.
On
$$n^2f(n)-(n-1)^2f(n-1)=f(n)\\(n^2-1)f(n)=(n-1)^2f(n-1)\\f(n)=\frac{(n-1)}{n+1}f(n-1)=\frac{n-1}{n+1}\cdot\frac{n-2}{n}f(n-2)=\cdots=\frac{(n-1)!}{(n+1)\cdot n\cdots4\cdot3}f(1)=\frac{2}{n(n+1)}f(1)\\f(2013)=\frac{2}{2013\cdot 2014}\cdot2014=\frac{2}{2013}$$
On
Calculating a few of the first $n$ by hand reveals a pattern:
$$f(n) = \frac{f(1)}{T_n},$$
where $T_n$ is the $n$th triangular number.
Proof by induction:
The relationship holds for $n=1$ since $T_1 = 1.$
We assume that
$$f(x) = \sum_{k=1}^n f(k) = \frac{2}{n(n+1)} f(1).$$
This implies that
$$\frac{2}{n(n+1)} f(1) = n^2 f(n).$$
Then,
$$f(x+1) = \sum_{k=1}^{n+1} f(k) = n^2 f(n) + f(n+1) = (n+1)^2 f(n+1),$$
and
$$f(n+1) = \frac{n}{n+2}f(n) = \frac{2f(1)}{n(n+1)}\frac{n}{n+2} = \frac{2f(1)}{(n+1)(n+2)} = \frac{f(1)}{T_{n+1}}.$$
To answer the question, then,
$$f(2013) = \frac{2 \cdot 2014}{2013 \cdot 2014} = \frac{2}{2013}.$$
Set $n=m,m+1$ one by one to find $f(m+1)$ in terms of $f(m)$
$$\sum_{r=1}^{m-1}f(r)=(m^2-1)f(m)$$
$$\{(m+1)^2-1\}f(m+1)=\sum_{r=1}^{m}f(r)=\sum_{r=1}^{m-1}f(r)+f(m)=(m^2-1)f(m)+f(m)$$
$$\iff m(m+2)f(m+1)=m^2f(m)\iff f(m+1)=\frac m{m+2}f(m)$$ for $m>0$