Difference between revisions of "1996 AJHSME Problems/Problem 16"
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Continuing the pattern, every group has a sum of <math>0</math>, and thus the entire sum is <math>0</math>, giving an answer of <math>\boxed{C}</math>. | Continuing the pattern, every group has a sum of <math>0</math>, and thus the entire sum is <math>0</math>, giving an answer of <math>\boxed{C}</math>. | ||
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| + | ==Solution 2== | ||
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| + | Let any term of the series be <math>t_n</math>. Realize that at every <math>n\equiv0 \pmod4</math>, the sum of the series is 0. And we know <math>1996\equiv0 \pmod4</math> so the solution is <math>\boxed{C}</math>. | ||
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| + | ~Golden_Phi | ||
==See Also== | ==See Also== | ||
Revision as of 21:10, 28 March 2024
Contents
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. And we know 
so the solution is .
~Golden_Phi
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
