Given a positive integer $ n$, for all positive integers $ a_1, a_2, \cdots, a_n$ that satisfy $ a_1 = 1$, $ a_{i + 1} \leq a_i + 1$, find $ \displaystyle \sum_{i = 1}^{n} a_1a_2 \cdots a_i$.
Problem
Source: China TST 2002 Quiz
Tags: algebra unsolved, algebra
31.01.2009 11:40
Shobber, do you mean this? Given a positive integer $ n$, find the sum $ \sum_{\begin{array}{c} a_1,a_2,\cdots ,a_n\text{ positive} \\ \text{integers satisfying} \\ a_1 = 1\text{ and} \\ a_{i + 1} \leq a_i + 1\text{ for all }i\end{array}}\sum_{i = 1}^{n} a_1a_2 \cdots a_i$. In this case, the sum equals $ 1$, $ 5$, $ 28$, $ 190$, $ 1594$ for $ n$ being $ 1$, $ 2$, $ 3$, $ 4$, $ 5$, respectively. I can't find this in Sloane's encyclopedia... darij
02.01.2016 16:37
First we can show by induction that for $i\in\{1,...,n\}$, we have $a_i\le i$. Thus $ \displaystyle \sum_{i = 1}^{n} a_1a_2 \cdots a_i\le 1!+2!+...+n!$, with equality if and only if $a_i=i$.
09.06.2017 05:46
As per request, here is a solution to this problem: