XV OM - I - Problem 10

Find the $ n $-th term of the sequence of numbers $ a_1, a_2, a_n, \ldots $, where $ a_1 = 1 $, $ a_2 = 3 $, $ a_3 = 6 $, and for every natural number $ k $
We can write equality (1) in the form






Let's denote $ a_{i+1} - a_i = r_i $ for $ i = 1, 2, 3, \ldots $; according to the above equality, $ r_{k+2} - 2r_{k+1} + r_k = 0 $, from which $ r_{k+2} - r_{k+1}= r_{k+1} - r_k $. 
The sequence of differences $ r_k $ is therefore an arithmetic progression. Since $ r_1 = a_2 - a_1 = 2 $ and $ r_2 = a_3 - a_2 = 3 $, the difference of this progression is $ r_2 - r_1 = 1 $. Hence, for every natural $ k $, we have $ r_k = r_1 + (k - 1) \cdot 1 = k + 1 $, thus






for every natural $ k $. From this, it easily follows (e.g., by induction) that $ a_n $ is the sum of natural numbers from $ 1 $ to $ n $:










