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Difference between revisions of "2007 UNCO Math Contest II Problems/Problem 1"

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== Solution ==
 
== Solution ==
  
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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 = 670004.</math> The second sequence includes <math>670</math> terms, so it equals: <cmath>670(-1)=-670.</cmath> Summing these two, we have: <cmath>670004-670 = \boxed{669334}.</cmath>
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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:

$1+ 2 - 3 + 4 + 5 - 6 + 7 + 8 - 9 +10 +11-12 +\cdots + 2005 + 2006 - 2007.$

Solution

We see that the 2nd and the 3rd terms make $-1$, the 5th and 6th terms make $-1$, and so on. Therefore, we are left with: \[1-1+4-1+7-1+10-1+\dots+2005-1.\] We have two sequences: \[1+4+7+10+\dots+2005\] and \[-1-1-1-1-\dots-1.\] The first sequence sums to $\dfrac{2006}{2}\cdot 669 = 671007.$ The second sequence includes $669$ terms, so it equals: \[669(-1)=-669.\] Summing these two, we have: \[671007-669 = \boxed{670338}.\]

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
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