Difference between revisions of "2002 AMC 10B Problems/Problem 23"
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Substituting <math>n=1</math> into <math>a_{m+n}=a_m+a_n+mn</math>: <math>a_{m+1}=a_m+a_{1}+m</math>. Since <math>a_1 = 1</math>, <math>a_{m+1}=a_m+m+1</math>. Therefore, <math>a_m = a_{m-1} + m, a_{m-1}=a_{m-2}+(m-1), a_{m-2} = a_{m-3} + (m-2)</math>, and so on until <math>a_2 = a_1 + 2</math>. Adding the Left Hand Sides of all of these equations gives <math>a_m + a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_2</math>; adding the Right Hand Sides of these equations gives <math>(a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_1) + (m + (m-1) + (m-2) + \cdots + 2)</math>. These two expressions must be equal; hence <math>a_m + a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_2 = (a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_1) + (m + (m-1) + (m-2) + \cdots + 2)</math> and <math>a_m = a_1 + (m + (m-1) + (m-2) + \cdots + 2)</math>. Substituting <math>a_1 = 1</math>: <math>a_m = 1 + (m + (m-1) + (m-2) + \cdots + 2) = 1+2+3+4+ \cdots +m = \frac{(m+1)(m)}{2}</math>. Thus we have a general formula for <math>a_m</math> and substituting <math>m=12</math>: <math>a_{12} = \frac{(13)(12)}{2} = (13)(6) = \boxed{\mathrm{(D) \ } 78}</math>. | Substituting <math>n=1</math> into <math>a_{m+n}=a_m+a_n+mn</math>: <math>a_{m+1}=a_m+a_{1}+m</math>. Since <math>a_1 = 1</math>, <math>a_{m+1}=a_m+m+1</math>. Therefore, <math>a_m = a_{m-1} + m, a_{m-1}=a_{m-2}+(m-1), a_{m-2} = a_{m-3} + (m-2)</math>, and so on until <math>a_2 = a_1 + 2</math>. Adding the Left Hand Sides of all of these equations gives <math>a_m + a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_2</math>; adding the Right Hand Sides of these equations gives <math>(a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_1) + (m + (m-1) + (m-2) + \cdots + 2)</math>. These two expressions must be equal; hence <math>a_m + a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_2 = (a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_1) + (m + (m-1) + (m-2) + \cdots + 2)</math> and <math>a_m = a_1 + (m + (m-1) + (m-2) + \cdots + 2)</math>. Substituting <math>a_1 = 1</math>: <math>a_m = 1 + (m + (m-1) + (m-2) + \cdots + 2) = 1+2+3+4+ \cdots +m = \frac{(m+1)(m)}{2}</math>. Thus we have a general formula for <math>a_m</math> and substituting <math>m=12</math>: <math>a_{12} = \frac{(13)(12)}{2} = (13)(6) = \boxed{\mathrm{(D) \ } 78}</math>. | ||
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+ | == Additional Comment== | ||
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+ | This is also the formula for the triangular numbers <math>T_n=1+2+3+...+n</math>, as seen in Solution 2 | ||
== See also == | == See also == | ||
{{AMC10 box|year=2002|ab=B|num-b=22|num-a=24}} | {{AMC10 box|year=2002|ab=B|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:18, 11 October 2015
Problem 23
Let be a sequence of integers such that and for all positive integers and Then is
Solution 1
First of all, write and in terms of
can be represented by in different ways.
Since both are equal to you can set them equal to each other.
Substitute the value of back into and substitute that into
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 : .
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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