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

(Solution)
(Solution)
Line 10: Line 10:
  
 
== Solution ==
 
== Solution ==
<math>\fbox{B}</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>.
It's easy to calculate that if <math>n</math> is even, <math>Sn</math> is negative <math>n</math> over <math>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) =9+17-25, which is </math>\fbox{B}$.
 
  
 
== See also ==
 
== See also ==

Revision as of 00:39, 18 September 2018

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

It's easy to calculate that if $n$ is even, $S_{n}$ is negative $n/2$. If $n$ is odd then $Sn$ is $(n+1)/2$. Therefore, we know $S_{17}+S_{33}+S_{50}$ =$9+17-25$, which is $\fbox{B}$.

See also

1968 AHSME (ProblemsAnswer KeyResources)
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. AMC logo.png