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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)==
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From inspection, we see that the sum of the sequence is <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.
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From inspection, we see that the sum of the sequence is <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.
  
 
==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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:45, 7 June 2021

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 1

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 $2008^{\text{th}}$ term of the sequence 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 $n^2$. 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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