1996 AJHSME Problems/Problem 16
Problem
Solution
Put the numbers in groups of 
:
The first group has a sum of 
.
The second group increases the two positive numbers on the end by 
, and decreases the two negative numbers in the middle by . Thus, the second group also has a sum of .
Continuing the pattern, every group has a sum of , and thus the entire sum is , giving an answer of 
.
Solution 2
Let any term of the series be 
. Realize that at every 
, the sum of the series is 0. For 
we know 
so the solution is .
~Golden_Phi
Solution 3
This solution is slightly more detailed than Solution 2, even though they are essentially the same.
Compute the sum of the first and last term, the second and second last term, the third and third last term. We get 
and so on. We see that each pair's sum alternates between 
and 
.We see that every two consecutive pair cancels out. For example 
gives a result of . There are a total of 
terms, there are 
pairs, and there are 
pairs which result in a sum of each and pairs which result in a sum of each. These all cancel out to a sum of , and thus the entire sum is , giving an answer of .
~blankbox
See Also
| 1996 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
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