- Thread starter
- #1

#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

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

(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

- Thread starter Albert
- Start date

- Thread starter
- #1

- Jan 25, 2013

- 1,225

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

(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

- Moderator
- #2

- Feb 7, 2012

- 2,725

$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

$-1,\ 0$ (two terms, with sum $-1$),

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

$1,\ 2,\ -3,\ 2,\ -3,\ -2,\ -1,\ 0$ (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 \(\displaystyle \Bigl|\sum_{n=1}^{2004}x_n\Bigr|\) 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 \(\displaystyle \Bigl|\sum_{n=1}^{2004}x_n\Bigr| = 10.\) As far as I can see, that is the smallest possible value for \(\displaystyle \Bigl|\sum_{n=1}^{2004}x_n\Bigr|.\)

- Thread starter
- #3

- Jan 25, 2013

- 1,225

thanks ,your answer is correct$-1,\ 0$ (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 \(\displaystyle \Bigl|\sum_{n=1}^{2004}x_n\Bigr|\) 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 \(\displaystyle \Bigl|\sum_{n=1}^{2004}x_n\Bigr| = 10.\) As far as I can see, that is the smallest possible value for \(\displaystyle \Bigl|\sum_{n=1}^{2004}x_n\Bigr|.\)