Bootin D.A.

Find in the sequence $2,6,12,20,30, \ldots$ the number standing a) at the 6th; b) at the 1994th position. Explain your answer.
$2=1 \cdot 2 ; 6=2 \cdot 3$

## Solution

We can notice that $2=1 \cdot 2, 6=2 \cdot 3, 12=3 \cdot 4$, and assume that the $n$-th term of the sequence is $n \cdot (n+1)$. Checking on the 4th ($20=4 \cdot 5$) and 5th ($30=5 \cdot 6$) terms of the sequence shows that we are correct. Therefore, the number at the 6th position is $6 \cdot 7=42$, and at the 1994th position, it is $1994 \cdot 1995=3978030$.

Of course, this is not a strict mathematical proof. For example, this method could "prove" that the number sixty is divisible by all numbers. Indeed, 60 is divisible by 1, by 2, by 3, by 4, by 5, by 6, etc. However, for solving the problem, it is only necessary to find a sufficiently simple rule that can generate such a sequence. The ability to see and feel the pattern (which was required in this problem) is just as important for a mathematician as the ability to reason strictly! If you find any other (but also "sufficiently simple") rule that gives the sequence $2, 6, 12, 20, 30$, please write to us (and such a solution would also be accepted on the olympiad!).

## Answer

a) 42; b) $1994 \cdot 1995=3978030$.

## Problem