Difference between revisions of "2008 AMC 10B Problems/Problem 13"
(→Solution 1) |
(→Solution 2 (Using Answer Choices)) |
||
Line 10: | Line 10: | ||
==Solution 2 (Using Answer Choices)== | ==Solution 2 (Using Answer Choices)== | ||
− | From inspection, we see that the sum of the sequence is | + | 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. |
==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}} | {{MAA Notice}} |
Revision as of 11:45, 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 numbers.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.