Difference between revisions of "2008 AMC 10B Problems/Problem 13"
I like pie (talk | contribs) (New page: ==Problem== {{problem}} ==Solution== {{solution}} ==See also== {{AMC10 box|year=2008|ab=B|num-b=12|num-a=14}}) |
(→Solution 1) |
||
(18 intermediate revisions by 10 users not shown) | |||
Line 1: | Line 1: | ||
==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 <math>2008^{\text{th}}</math> term of the sequence? |
− | ==Solution== | + | <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 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 <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. | ||
==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 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.