1999 AHSME Problems/Problem 1
Contents
Problem
Solution
Solution 1
If we group consecutive terms together, we get , and since there are 49 pairs of terms the answer is .
Solution 2
( Similar to Solution 1 ) If we rearranged the terms, we get then , and since there are 49 pairs of terms and the in the beginning the answer is .
Solution 3
Let .
Therefore,
We add:
Solution 4
We proceed with addition, 1 -2 + 3 -4.... Once done we find
See also
1999 AHSME (Problems • Answer Key • Resources) | ||
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Followed by Problem 2 | |
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