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Difference between revisions of "2006 SMT/General Problems/Problem 10"

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==Solution==
 
==Solution==
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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 = \boxed{4024036}</math>
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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>

Revision as of 17:09, 13 January 2020

Solution

The sum of the first n odd integers is $n^2$. This comes from the fact that $(n+1)^2-n^2 = 2n+1$ (Taking a sum of this equation beginning with $n = 0$ will yield the desired result as the LHS will telescope). Therefore, the sum of the first 2006 odd integers is $2006^2$. The answer we are looking for is $\sqrt{2006^2} = \boxed{2006}$

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