2008 AMC 10B Problems/Problem 13: Difference between revisions
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==Solution== | ==Solution== | ||
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 2008th term 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 numbers. | Note that <math>n^2</math> is the sum of the first n odd numbers. | ||
Revision as of 11:25, 24 October 2015
Problem
For each positive integer
, the mean of the first
terms of a sequence is
. What is the
term of the sequence?
Solution
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 2008th term is
Note that
is the sum of the first n odd numbers.
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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