Difference between revisions of "2014 UNCO Math Contest II Problems/Problem 2"
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== Solution == | == Solution == | ||
+ | It is evident that we are being asked to find the sum of the sequence <math>1-2+3-4+...-288+289</math>. | ||
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+ | A straight-forward approach is to find the sum of all the odd numbers from 1 to 289, inclusive, and subtract the sum of all the even numbers from 2 to 288, inclusive. Doing so would give you <math>145^2 - (144^2 + 144) = (145 - 144)(145 + 144) - 144 = (1)(145 + 144) - 144 = \boxed{145}</math>. | ||
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+ | However, a faster method would be to see that you can group the terms in the sequence <math>1-2+3-4+...-288+289</math> into <math>1 + (-2 + 3) + (-4 + 5) +...+ (-288 + 289)</math>. Using this, we can reduce many pairs of positive and negative numbers to <math>1</math>s. Since there are as many <math>1</math>s as there are odd numbers between 1 and 289, inclusive, and there are 145 odd numbers, the answer is <math>(1)(145) = \boxed{145}</math>. | ||
== See also == | == See also == |
Latest revision as of 10:28, 24 December 2019
Problem
Define the Cheshire Cat function by
Find the sum
Solution
It is evident that we are being asked to find the sum of the sequence .
A straight-forward approach is to find the sum of all the odd numbers from 1 to 289, inclusive, and subtract the sum of all the even numbers from 2 to 288, inclusive. Doing so would give you .
However, a faster method would be to see that you can group the terms in the sequence into . Using this, we can reduce many pairs of positive and negative numbers to s. Since there are as many s as there are odd numbers between 1 and 289, inclusive, and there are 145 odd numbers, the answer is .
See also
2014 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
All UNCO Math Contest Problems and Solutions |