Difference between revisions of "1968 AHSME Problems/Problem 27"

(Solution)
(Solution)
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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>S_{n}</math> is <math>(n+1)/2</math>.
+
If <math>n</math> is even, <math>S_{n}</math> is negative <math>n/2</math>. If <math>n</math> is odd, then <math>S_{n}</math> is <math>(n+1)/2</math>.
 +
(These can be found using simple calculations.)
 
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
 
Solution By FrostFox
(Quite a simple problem once you think about it.)
 
  
 
== See also ==
 
== See also ==

Revision as of 01:17, 27 January 2019

Problem

Let $S_n=1-2+3-4+\cdots +(-1)^{n-1}n$, where $n=1,2,\cdots$. Then $S_{17}+S_{33}+S_{50}$ equals:

$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } -1\quad \text{(E) } -2$

Solution

If $n$ is even, $S_{n}$ is negative $n/2$. If $n$ is odd, then $S_{n}$ is $(n+1)/2$. (These can be found using simple calculations.) Therefore, we know $S_{17}+S_{33}+S_{50}$ =$9+17-25$, which is $\fbox{B}$. Solution By FrostFox

See also

1968 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 26
Followed by
Problem 28
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