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 = 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> | + | 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> |
== See Also == | == See Also == |
Revision as of 21:21, 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:
We have two sequences:
and
The first sequence sums to
The second sequence includes
terms, so it equals:
Summing these two, we have:
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 |