1984 AIME Problems/Problem 1
Contents
Problem
Find the value of if , , is an arithmetic progression with common difference 1, and .
Solution
Solution 1
One approach to this problem is to apply the formula for the sum of an arithmetic series in order to find the value of , then use that to calculate and sum another arithmetic series to get our answer.
A somewhat quicker method is to do the following: for each , we have . We can substitute this into our given equation to get . The left-hand side of this equation is simply , so our desired value is .
Solution 2
If is the first term, then can be rewritten as:
Our desired value is so this is:
which is . So, from the first equation, we know . So, the final answer is:
.
Solution 3
A better approach to this problem is to notice that from that each element with an odd subscript is 1 from each element with an even subscript. Thus, we note that the sum of the odd elements must be . Thus, if we want to find the sum of all of the even elements we simply add common differences to this giving us .
Or, since the sum of the odd elements if 44, then the sum of the even terms must be .
Solution 4
We want to find the value of , which can be rewritten as . We can split into two parts: and Note that each term in the second expression is greater than the corresponding term, so, letting the first equation be equal to , we get . Calculating by sheer multiplication is not difficult, but you can also do . We want to find the value of . Since , we find . .\\ - PhunsukhWangdu
See also
1984 AIME (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |