2002 AMC 10B Problems/Problem 23
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
We can set equal to , so we can say that
We set , we get .
We set m, we get .
Solving for is easy, just direct substitution.
Substituting, we get
Thus, the answer is .
~ euler123
Solution 5
Note that the sequence of triangular numbers satisfies these conditions. It is immediately obvious that it satisfies , and can be visually proven with the diagram below.
This means that we can use the triangular number formula , so the answer is .
Solution 6
Follow solution 4 with plugging in m=n, but for the last part since since the subscript is 2n, and we aren't given anything that has odd numbers ( without substitution into the original equation), we need to find two numbers that add up to 12, which result from the multiplication of two even numbers. We realize that 4+8 works, since 4=2x2 and 8=4x2. We then proceed in substituting n=2 and n=4 into the equation, then add to obtain 78. ~Charmainema07292010
Video Solution
https://www.youtube.com/watch?v=zraGzYAh0uM ~David
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 | ||
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