Hellenic Mathematical Olympiad

We observe that: $$a_1=2,a_2=\frac{3}{2}\cdot a_1=3\cdot 2,a_3=\frac{4}{2}\cdot (a_1+a_2)=\frac{4}{2}\cdot 4\cdot 2=4\cdot 2^2,$$ $$a_4=\frac{5}{3}\cdot (a_1+a_2+a_3)=\frac{5}{3}\cdot 24=5\cdot 2^3,$$ $$a_5=\frac{6}{4}\cdot (a_1+a_2+a_3+a_4)=\frac{6}{4}\cdot 64=6\cdot 2^4.$$ We are going to use induction. Let $a_n=(n+1)\cdot 2^{n-1}$, for $n=1,2,3,...,k$. We will prove that the same formula is valid for $n=k+1$, i.e.: $a_{k+1}=(k+2)\cdot 2^k$. $$a_{k+1}=\frac{k+2}{k}(a_1+a_2+a_3+\cdots +a_k)=\frac{k+2}{k}\left(2+3\cdot 2^1+4\cdot 2^2+\cdots +(k+1)\cdot 2^{k-1}\right).$$ Hence: $$a_{k+1}=\frac{k+2}{k}\left(2\cdot 2^0+3\cdot 2^1+4\cdot 2^2+\cdots +(k+1)\cdot 2^{k-1}\right).(1)$$ Multiplying both parts of relation (1) by 2 we get: $$2a_{k+1}=\frac{k+2}{k}\left(2^1+3\cdot 2^2+4\cdot 2^3+\cdots +(k+1)\cdot 2^k\right),(2)$$ And then from (1) and (2) we find: $$\begin{gathered} a_{k+1}=\frac{k+2}{k}\left(-2-2^1-2^2-2^3-\cdots -2^{k-1}+(k+1)\cdot 2^k\right) \\ {a}_{k+1}=\frac{k+2}{k}\left(-1-\frac{1-2^k}{1-2}+(k+1)\cdot 2^k\right)=\frac{k+2}{k}\left(-2^k+(k+1)\cdot 2^k\right)=(k+2)\cdot 2^k. \end{gathered}$$ Therefore we have $a_{2013}=2014\cdot 2^{2012}$

---

Alternative solution.

From equations $$a_n=\left(\frac{n+1}{n-1}\right)(a_1+a_2+\cdots +a_{n-1}),n\ge 2,(3)$$ $$a_{n+1}=\left(\frac{n+2}{n}\right)(a_1+a_2+\cdots +a_n),n\ge 1,(4)$$ we find $$a_1+a_2+\cdots +a_{n-1}=\left(\frac{n-1}{n+1}\right)a_n,n\ge 2(5)$$ $$a_1+a_2+\cdots +a_n=\left(\frac{n}{n+2}\right)a_{n+1},n\ge 1(6)$$ And from (5) and (6) we get: $$a_n=\left(\frac{n}{n+2}\right)a_{n+1}-\left(\frac{n-1}{n+1}\right)a_n\Rightarrow a_{n+1}=\left(\frac{2(n+2)}{n+1}\right)a_n,n\ge 1(7)$$ Hence $$\begin{gathered} a_n=\left(\frac{2(n+1)}{n}\right)a_{n-1}=\left(\frac{2(n+1)}{n}\right)\left(\frac{2n}{n-1}\right)a_{n-2}=\dots \\ =\left(\frac{2(n+1)}{n}\right)\left(\frac{2n}{n-1}\right)\dots \left(\frac{2\cdot 4}{3}\cdot \frac{2\cdot 3}{2}\right)a_1 \\ =(n+1)\cdot 2^{n-2}\cdot a_1=(n+1)\cdot 2^{n-1},\text{ since }{a}_1=2. \end{gathered}$$