MHB Finding the Min Sum of a Sequence of Numbers in $Z$

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To find the minimum sum of a sequence of integers where each term's absolute value is one more than the previous term, starting from zero, sequences can be constructed that return to zero while minimizing the overall sum. It is established that for any integer n, sequences of 2n terms can yield a negative sum of -n. A proposed solution involves using a specific sequence where the first 1960 terms yield a sum of -980, and the last 44 terms contribute a sum of 990, resulting in a total of 10. This value of 10 is identified as the smallest possible absolute value for the sum of the sequence.
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$x_0,x_1,-----,x_{2004} \in Z , \, x_0=0, \mid x_n \mid =\mid x_{n-1}+1\mid $

$for, \,\, 1 \leq n \leq 2004$

(1) $find :\,\, min\mid x_1+x_2+x_3+ ------+x_{2004}\mid $

(2) get a set of numbers $ x_1,x_2,-----x_{2004} $ satisfying your answer
 
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Albert said:
$x_0,x_1,-----,x_{2004} \in Z , \, x_0=0, \mid x_n \mid =\mid x_{n-1}+1\mid $

$for, \,\, 1 \leq n \leq 2004$

(1) $find :\,\, \min\mid x_1+x_2+x_3+ ------+x_{2004}\mid $

(2) get a set of numbers $ x_1,x_2,-----x_{2004} $ satisfying your answer
[sp]Starting from $0$, you can get sequences of terms ending with another zero, for example
$-1,\ 0$ (two terms, with sum $-1$),

$1,\ 2,\ -3,\ 2,\ -3,\ -2,\ -1,\ 0$ (eight terms, with sum $-4$).​

In fact, for any $n$, starting from $0$ you can find sequences of $2n$ terms that bring you back to $0$, but the sum of the terms in the sequence will always be $-n$. So to minimise $$\Bigl|\sum_{n=1}^{2004}x_n\Bigr|$$ you need to build up a negative sum from such sequences, and then end with a run of positive terms to cancel out as much as possible of the negative sum. For example, suppose that $$x_n = \begin{cases}-1&(n \text{ odd, }1\leqslant n\leqslant 1959), \\ 0&(n \text{ even, }2\leqslant n\leqslant 1960), \\ n-1960&(1961\leqslant n\leqslant 2004).\end{cases}$$ Then the sum of the first 1960 terms is $-980$, the sum of the remaining terms is $1+2+3+\ldots+44 = 990$ and therefore $$\Bigl|\sum_{n=1}^{2004}x_n\Bigr| = 10.$$ As far as I can see, that is the smallest possible value for $$\Bigl|\sum_{n=1}^{2004}x_n\Bigr|.$$[/sp]
 
Opalg said:
[sp]Starting from $0$, you can get sequences of terms ending with another zero, for example
$-1,\ 0$ (two terms, with sum $-1$),

$1,\ 2,\ -3,\ 2,\ -3,\ -2,\ -1,\ 0$ (eight terms, with sum $-4$).​

In fact, for any $n$, starting from $0$ you can find sequences of $2n$ terms that bring you back to $0$, but the sum of the terms in the sequence will always be $-n$. So to minimise $$\Bigl|\sum_{n=1}^{2004}x_n\Bigr|$$ you need to build up a negative sum from such sequences, and then end with a run of positive terms to cancel out as much as possible of the negative sum. For example, suppose that $$x_n = \begin{cases}-1&(n \text{ odd, }1\leqslant n\leqslant 1959), \\ 0&(n \text{ even, }2\leqslant n\leqslant 1960), \\ n-1960&(1961\leqslant n\leqslant 2004).\end{cases}$$ Then the sum of the first 1960 terms is $-980$, the sum of the remaining terms is $1+2+3+\ldots+44 = 990$ and therefore $$\Bigl|\sum_{n=1}^{2004}x_n\Bigr| = 10.$$ As far as I can see, that is the smallest possible value for $$\Bigl|\sum_{n=1}^{2004}x_n\Bigr|.$$[/sp]
thanks ,your answer is correct :)
 
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