Value of function

Let $f$ be a function defined as $f(1)=2$ and for every positive integer $n>1$ holds:

$$f(1)+f(2)+f(3)+\cdots +f(n) = n^2 f(n)$$

Evaluate the value $f(2013)$.

Source: CRUX

Solution



 For $n>1$ we have that:

$$n^2 f(n)= \left [f(1)+f(2)+f(3)+\cdots + f(n-1)\right ]+ f(n) = (n-1) ^2f(n-1)+f(n)$$

From the initial and the last we deduce that $\displaystyle  f(n) = \frac { n-1}{n+1} f(n-1)$.

Hence we can now get a recurrence relation:

$$\begin{aligned}
f(n) &=\frac { n-1}{n+1} \cdot \frac { n-2}{n}\cdot  f(n-2) \\
 &=\cdots \\
 &=\frac { n-1}{n+1} \cdot  \frac { n-2}{n}\cdots \frac { 2}{4}\cdot \frac { 1}{3} \cdot f(1)  \\
 &= \frac { 4\cdot 1}{(n+1)n} 
\end{aligned}$$

The last for $n=2013$ gives $\displaystyle f(2013)= \frac{2}{1007 \cdot 2013}$ completing the exercise.