Difference between revisions of "1968 AHSME Problems/Problem 27"
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== Solution == | == Solution == | ||
| − | It's easy to calculate that if <math>n</math> is even, <math>S_{n}</math> is negative <math>n/2</math>. If <math>n</math> is odd then <math>Sn</math> is <math>(n+1)/2</math>. | + | It's easy to calculate that if <math>n</math> is even, <math>S_{n}</math> is negative <math>n/2</math>. If <math>n</math> is odd, then <math>Sn</math> is <math>(n+1)/2</math>. |
Therefore, we know <math>S_{17}+S_{33}+S_{50}</math> =<math>9+17-25</math>, which is <math>\fbox{B}</math>. | Therefore, we know <math>S_{17}+S_{33}+S_{50}</math> =<math>9+17-25</math>, which is <math>\fbox{B}</math>. | ||
| + | Solution By FrostFox | ||
== See also == | == See also == | ||
Revision as of 23:39, 17 September 2018
Problem
Let 
, where 
. Then 
equals:
Solution
It's easy to calculate that if 
is even, 
is negative 
. If is odd, then 
is 
.
Therefore, we know =
, which is 
.
Solution By FrostFox
See also
| 1968 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 26 |
Followed by Problem 28 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
