Difference between revisions of "2006 SMT/General Problems/Problem 10"
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| − | The sum of the first n odd integers is <math>n^2</math>. This comes from the fact that <math>(n+1)^2-n^2 = 2n+1</math> (Taking a sum of this equation beginning with <math>n = 0</math> will yield the desired result as the LHS will telescope). Therefore, the sum of the first 2006 odd integers is <math>2006^2</math>. The answer we are looking for is <math>\sqrt{2006^2} = \boxed{2006}</math> | + | The sum of the first n positive odd integers is <math>n^2</math>. This comes from the fact that <math>(n+1)^2-n^2 = 2n+1</math> (Taking a sum of this equation beginning with <math>n = 0</math> will yield the desired result as the LHS will telescope). Therefore, the sum of the first 2006 odd integers is <math>2006^2</math>. The answer we are looking for is <math>\sqrt{2006^2} = \boxed{2006}</math> |
Revision as of 17:10, 13 January 2020
Solution
The sum of the first n positive odd integers is 
. This comes from the fact that 
(Taking a sum of this equation beginning with 
will yield the desired result as the LHS will telescope). Therefore, the sum of the first 2006 odd integers is 
. The answer we are looking for is 
