2013
Problem - 3165
Given the sequence $\{a_n\}$ satisfies $a_n+a_m=a_{n+m}$ for any positive integers $n$ and $m$. Suppose $a_1=\frac{1}{2013}$. Find the sum of its first $2013$ terms.
$$
\begin{array}{lclclcl}
a_2 &=& a_1 & +& a_1 & = &2a_1\\
a_3 &=& a_2 &+ &a_1 & = &3a_1\\
\cdots \\
a_{2013} & =& a_{2012}&+&a_1 &= &2013 a_1
\end{array}
$$
\begin{align*}
\therefore\qquad &a_1+a_2+\cdots + a_{2013}\\
= & a_1 + 2a_2 + \cdots 2013a_1\\
= &(1+2+\cdots + 2013)a_1\\
= & \frac{(1+2013)\times 2013}{2}\times\frac{1}{2013}\\
= & \boxed{1007}
\end{align*}