2008 AMC 10B Problems/Problem 13

Revision as of 12:36, 27 August 2008 by Abacadaea (talk | contribs) (Solution)

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 $2008^2-2007^2=(2008+2007)(2008-2007)=4015\Rightarrow \boxed{B}$

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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All AMC 10 Problems and Solutions
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