Difference between revisions of "2002 AMC 10B Problems/Problem 23"
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We can literally just plug stuff in. No prerequisite is actually said in the sequence. Since <math>a_{m+n} = a_m+a_n +mn</math>, we know <math>a_2=a_1+a_1+1\cdot1=1+1+1=3</math>. After this, we can use <math>a_2</math> to find <math>a_4</math>. <math>a_4=a_2+a_2+2\cdot 2 = 3+3+4 = 10</math>. Now, we can use <math>a_2</math> and <math>a_4</math> to find <math>a_6</math>, or <math>a_6=a_4+a_2+4\cdot 2 = 10+3+8=21</math>. Lastly, we can use <math>a_6</math> to find <math>a_{12}</math>. <math>a_{12} = a_6+a_6+6\cdot 6 = 21+21+36= \boxed{(D) 78}</math> | We can literally just plug stuff in. No prerequisite is actually said in the sequence. Since <math>a_{m+n} = a_m+a_n +mn</math>, we know <math>a_2=a_1+a_1+1\cdot1=1+1+1=3</math>. After this, we can use <math>a_2</math> to find <math>a_4</math>. <math>a_4=a_2+a_2+2\cdot 2 = 3+3+4 = 10</math>. Now, we can use <math>a_2</math> and <math>a_4</math> to find <math>a_6</math>, or <math>a_6=a_4+a_2+4\cdot 2 = 10+3+8=21</math>. Lastly, we can use <math>a_6</math> to find <math>a_{12}</math>. <math>a_{12} = a_6+a_6+6\cdot 6 = 21+21+36= \boxed{(D) 78}</math> | ||
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+ | ==Solution 4== | ||
== Additional Comment== | == Additional Comment== |
Revision as of 16:00, 27 December 2019
Contents
Problem 23
Let be a sequence of integers such that and for all positive integers and Then is
Solution 1
When , . Hence, Adding these equations up, we have that
~AopsUser101
Solution 2
Substituting into : . Since , . Therefore, , and so on until . Adding the Left Hand Sides of all of these equations gives ; adding the Right Hand Sides of these equations gives . These two expressions must be equal; hence and . Substituting : . Thus we have a general formula for and substituting : .
Solution 3
We can literally just plug stuff in. No prerequisite is actually said in the sequence. Since , we know . After this, we can use to find . . Now, we can use and to find , or . Lastly, we can use to find .
Solution 4
Additional Comment
This is also the formula for the triangular numbers , as seen in Solution 2
See also
2002 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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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