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Difference between revisions of "2008 AMC 10B Problems/Problem 13"
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==Solution 2(Using Answer Choices)==
 
==Solution 2(Using Answer Choices)==
−
From inspection, we see that the sum of the sequence is basically <math>n^2</math>. We also notice that <math>n^2</math> Is the sum of the first n ODD integers. Because 4015 is the only odd integer, <math>\boxed{B}</math> is the answer.
+
From inspection, we see that the sum of the sequence is basically <math>n^2</math>. We also notice that <math>n^2</math> Is the sum of the first <math>n</math> ODD integers. Because <math>4015</math> is the only odd integer, <math>\boxed{B}</math> is the answer.
  
  

Revision as of 09:15, 16 May 2017

Problem

For each positive integer $n$, the mean of the first $n$ terms of a sequence is $n$. What is the $2008^{\text{th}}$ 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 $2008^2-2007^2=(2008+2007)(2008-2007)=4015\Rightarrow \boxed{\text{(B)}}$

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


Solution 2(Using Answer Choices)

From inspection, we see that the sum of the sequence is basically $n^2$. We also notice that $n^2$ Is the sum of the first $n$ ODD integers. Because $4015$ is the only odd integer, $\boxed{B}$ is the answer.


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

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