1968 AHSME Problems/Problem 27

Revision as of 02:17, 27 January 2019 by Frostfox (talk | contribs) (Solution)

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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