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.: $a_{k+1} = (k+2) \cdot 2^k$.
$$a_{k+1} = \frac{k+2}{k} (a_1 + a_2 + a_3 + \dots + a_k) = \frac{k+2}{k} \left( 2 + 3 \cdot 2^1 + 4 \cdot 2^2 + \dots + (k+1) \cdot 2^{k-1} \right).$$
Hence:
$$a_{k+1} = \frac{k+2}{k} \left( 2 \cdot 2^0 + 3 \cdot 2^1 + 4 \cdot 2^2 + \dots + (k+1) \cdot 2^{k-1} \right). \quad (1)$$
Multiplying both parts of relation (1) by 2 we get:
$$2a_{k+1} = \frac{k+2}{k} \left( 2^1 + 3 \cdot 2^2 + 4 \cdot 2^3 + \dots + (k+1) \cdot 2^k \right), \quad (2)$$
And then from (1) and (2) we find:
$$a_{k+1} = \frac{k+2}{k} \left( -2 - 2^1 - 2^2 - 2^3 - \dots - 2^{k-1} + (k+1) \cdot 2^k \right) \\ a_{k+1} = \frac{k+2}{k} \left( -1 - \frac{1-2^k}{1-2} + (k+1) \cdot 2^k \right) = \frac{k+2}{k} \left( -2^k + (k+1) \cdot 2^k \right) = (k+2) \cdot 2^k.$$
Therefore we have $a_{2013} = 2014 \cdot 2^{2012}$

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