Difference between revisions of "2008 AMC 10B Problems/Problem 13"
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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> | 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 | + | Note that <math>n^2</math> is the sum of the first n odd integers. |
==Solution 2 (Using Answer Choices)== | ==Solution 2 (Using Answer Choices)== |
Revision as of 12:45, 7 June 2021
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.
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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