Difference between revisions of "2008 AMC 10B Problems/Problem 13"
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==Problem== | ==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 | + | 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> | <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 | + | 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== | ==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}} |
Latest revision as of 22:20, 30 August 2022
Contents
Problem
For each positive integer , the mean of the first terms of a sequence is . What is the term of the sequence?
Solution 1
Since the mean of the first terms is , the sum of the first terms is . Thus, the sum of the first terms is and the sum of the first terms is . Hence, the term of the sequence is
Note that 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 . We also notice that is the sum of the first odd integers. Because is the only odd integer, is the answer.
Solution 3 (Basically Solution Two Just More Rigorous)
Let be the terms of the sequence. We know , so we must have . The sum of consecutive odd numbers down to is a perfect square, if you don't believe me, try drawing squares with the sum, so , so the answer is .
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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