Difference between revisions of "2007 UNCO Math Contest II Problems/Problem 1"
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== Solution == | == Solution == | ||
| + | We see that the 2nd and the 3rd terms make <math>-1</math>, the 5th and 6th terms make <math>-1</math>, and so on. Therefore, we are left with: <cmath>1-1+4-1+7-1+10-1+\dots+2005-1.</cmath> We have two sequences: <cmath>1+4+7+10+\dots+2005</cmath> and <cmath>-1-1-1-1-\dots-1.</cmath> The first sequence sums to <math>\dfrac{2006}{2}\cdot 669 = 671007.</math> The second sequence includes <math>669</math> terms, so it equals: <cmath>669(-1)=-669.</cmath> Summing these two, we have: <cmath>671007-669 = \boxed{670338}.</cmath> | ||
== See Also == | == See Also == | ||
Latest revision as of 21:37, 4 February 2016
Problem
Express the following sum as a whole number:
Solution
We see that the 2nd and the 3rd terms make 
, the 5th and 6th terms make , and so on. Therefore, we are left with: ![\[1-1+4-1+7-1+10-1+\dots+2005-1.\]](http://latex.artofproblemsolving.com/b/5/1/b517da11c5c247dac8e87c669b654dd62aec8b6d.png)
We have two sequences: ![\[1+4+7+10+\dots+2005\]](http://latex.artofproblemsolving.com/c/6/9/c692889a6ee7616801e88f0cdc92c90b27f3d2d6.png)
and ![\[-1-1-1-1-\dots-1.\]](http://latex.artofproblemsolving.com/8/2/c/82c636804fab8fe5de0641b0bdb585d96d821891.png)
The first sequence sums to 
The second sequence includes 
terms, so it equals: ![\[669(-1)=-669.\]](http://latex.artofproblemsolving.com/d/6/3/d63aa2a5c58212d4b62f02ff5a2c7b6cf79e7f68.png)
Summing these two, we have: ![\[671007-669 = \boxed{670338}.\]](http://latex.artofproblemsolving.com/c/8/4/c84e97b11bb33c878829db80c81b34018ff5933a.png)
See Also
| 2007 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
