Difference between revisions of "2008 AMC 10B Problems/Problem 13"
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==Solution 2(Using Answer Choices)== | ==Solution 2(Using Answer Choices)== | ||
− | From inspection, we see that the sum of the sequence is basically <math>n^2</math>. We also notice that <math>n^2</math> Is the sum of the first n ODD integers. Because 4015 is the only odd integer, <math>\boxed{ | + | From inspection, we see that the sum of the sequence is basically <math>n^2</math>. We also notice that <math>n^2</math> Is the sum of the first n ODD integers. Because 4015 is the only odd integer, <math>\boxed{B}</math> is the answer. |
Revision as of 10:13, 16 May 2017
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.
Solution 2(Using Answer Choices)
From inspection, we see that the sum of the sequence is basically . We also notice that
Is the sum of the first n ODD integers. Because 4015 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.