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

Latest revision as of 22:20, 30 August 2022

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 integers.

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.

Solution 3 (Basically Solution Two Just More Rigorous)

Let $a_1, a_2, a_3, \cdots, a_n$ be the terms of the sequence. We know $\frac{a_1 + a_2 + a_3 + \cdots + a_n}{n} = n$ , so we must have $a_1 + a_2 + a_3 + \cdots + a_n = n^2$ . The sum of consecutive odd numbers down to $1$ is a perfect square, if you don't believe me, try drawing squares with the sum, so $a_1 = 1, a_2 = 3, a_3 = 5, \cdots , a_n = 2(n-1) + 1$ , so the answer is $a_{2008} = 2(2007) + 1 = \boxed{\text{B}}$ .

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
 ==Problem==
-For each positive integer <math>n</math>, the mean of the first <math>n</math> terms of a sequence is <math>n</math>. What is the 2008th term of the sequence?
+For each positive integer <math>n</math>, the mean of the first <math>n</math> terms of a sequence is <math>n</math>. What is the <math>2008^{\text{th}}</math> term of the sequence?
 <math>\mathrm{(A)}\ {{{2008}}} \qquad \mathrm{(B)}\ {{{4015}}} \qquad \mathrm{(C)}\ {{{4016}}} \qquad \mathrm{(D)}\ {{{4,030,056}}} \qquad \mathrm{(E)}\ {{{4,032,064}}}</math>
-==Solution==
+==Solution 1==
-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 <math>2008^{\text{th}}</math> term of the sequence is <math>2008^2-2007^2=(2008+2007)(2008-2007)=4015\Rightarrow \boxed{\text{(B)}}</math>
+Note that <math>n^2</math> is the sum of the first n odd integers.
+==Solution 2 (Using Answer Choices)==
+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.
+==Solution 3 (Basically Solution Two Just More Rigorous)==
+Let <math>a_1, a_2, a_3, \cdots, a_n</math> be the terms of the sequence. We know <math>\frac{a_1 + a_2 + a_3 + \cdots + a_n}{n} = n</math>, so we must have <math>a_1 + a_2 + a_3 + \cdots + a_n = n^2</math>. The sum of consecutive odd numbers down to <math>1</math> is a perfect square, if you don't believe me, try drawing squares with the sum, so <math>a_1 = 1, a_2 = 3, a_3 = 5, \cdots , a_n = 2(n-1) + 1</math>, so the answer is <math>a_{2008} = 2(2007) + 1 = \boxed{\text{B}}</math>.
 ==See also==
 {{AMC10 box|year=2008|ab=B|num-b=12|num-a=14}}
+{{MAA Notice}}

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