Let $ (a_n)^{\infty}_{n=1}$ be a sequence of integers with $ a_{n} < a_{n+1}, \quad \forall n \geq 1.$ For all quadruple $ (i,j,k,l)$ of indices such that $ 1 \leq i < j \leq k < l$ and $ i + l = j + k$ we have the inequality $ a_{i} + a_{l} > a_{j} + a_{k}.$ Determine the least possible value of $ a_{2008}.$
Problem
Source: MEMO 2008, Single, Problem 1
Tags: inequalities, algebra unsolved, algebra
11.09.2008 02:44
we have$ a_{i} + a_{i - 2}\geq 2a_{i - 1} + 1$ so we have $ a_{i} - a_{i - 1}\geq a_{i - 1} - a_{i - 2} + 1\geq.....a_{2} - a_{1} + (i - 2)\geq i - 1$ so we have $ a_{i}\geq \frac{i^2 - i + 2}{2}$. so min of $ a_{i}$ is $ \frac {i^2 - i + 2}{2}$ .and its easy to check the condition because $ i^2 + l^2 > j^2 + k^2$ .so the min of$ a_{2008}$ is 2015029.
02.10.2008 23:44
Note: It says in the question that $ 1 \leq i < j \leq k < l$ and not $ 1 \leq i < j < k < l$. The answer to the problem is $ 2^{2007}$.
31.01.2009 18:11
I may be wrong but i think that the sequence a(n) = n*(n+1)/2 works. So a(2008) = 1004*2009 = 2017036
16.05.2017 16:51
sumita is correct.
16.05.2017 18:22
I think 673015