Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
(→‎Solution)
Line 1: Line 1:
 
==Solution==
 
==Solution==
−
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>
+
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 18:09, 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 $\sqrt{2006^2} = \boxed{2006}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_SMT/General_Problems/Problem_10&oldid=114687"