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
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

Difference between revisions of "2008 AMC 10B Problems/Problem 13"

(→‎Solution)
m (→‎Solution)
Line 6: Line 6:
 
==Solution==
 
==Solution==
 
Since the mean of the first <math>n</math> terms is <math>n</math>, the sum of the first <math>n</math> terms is <math>n^2</math>. Thus, the sum of the first <math>2007</math> terms is <math>2007^2</math> and the sum of the first <math>2008</math> terms is <math>2008^2</math>. Hence, the 2008th term is <math>2008^2-2007^2=(2008+2007)(2008-2007)=4015\Rightarrow \boxed{B}</math>
 
Since the mean of the first <math>n</math> terms is <math>n</math>, the sum of the first <math>n</math> terms is <math>n^2</math>. Thus, the sum of the first <math>2007</math> terms is <math>2007^2</math> and the sum of the first <math>2008</math> terms is <math>2008^2</math>. Hence, the 2008th term is <math>2008^2-2007^2=(2008+2007)(2008-2007)=4015\Rightarrow \boxed{B}</math>
 +
 +
Note that <math>n^2</math> is the sum of the first n odd numbers.
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2008|ab=B|num-b=12|num-a=14}}
 
{{AMC10 box|year=2008|ab=B|num-b=12|num-a=14}}

Revision as of 11:57, 14 February 2009

Problem

For each positive integer $n$, the mean of the first $n$ terms of a sequence is $n$. What is the 2008th term of the sequence?

$\mathrm{(A)}\ {{{2008}}} \qquad \mathrm{(B)}\ {{{4015}}} \qquad \mathrm{(C)}\ {{{4016}}} \qquad \mathrm{(D)}\ {{{4,030,056}}} \qquad \mathrm{(E)}\ {{{4,032,064}}}$

Solution

Since the mean of the first $n$ terms is $n$, the sum of the first $n$ terms is $n^2$. Thus, the sum of the first $2007$ terms is $2007^2$ and the sum of the first $2008$ terms is $2008^2$. Hence, the 2008th term is $2008^2-2007^2=(2008+2007)(2008-2007)=4015\Rightarrow \boxed{B}$

Note that $n^2$ is the sum of the first n odd numbers.

See also

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10B_Problems/Problem_13&oldid=30259"