<s>[INST] Find value of f(2013)? 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.
 [/INST] 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,
$$f(2013) = \frac{2 \cdot 2014}{2013 \cdot 2014} = \frac{2}{2013}.$$
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