2002 AMC 10B Problems/Problem 23
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 .
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 | |
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