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Difference between revisions of "1988 AHSME Problems/Problem 14"

(Created page with "==Problem== For any real number a and positive integer k, define <math>{a \choose k} = \frac{a(a-1)(a-2)\cdots(a-(k-1))}{k(k-1)(k-2)\cdots(2)(1)}</math> What is <math>{-\frac...")
 
m (Fixed a typo)
 
(One intermediate revision by one other user not shown)
Line 17: Line 17:
 
==Solution==
 
==Solution==
  
 +
We expand both the numerator and the denominator.
  
 +
<cmath>\begin{align*}
 +
\binom{-\frac{1}{2}}{100}\div\binom{\frac{1}{2}}{100}
 +
&= \frac{
 +
  \dfrac{
 +
    (-\frac{1}{2})
 +
    (-\frac{1}{2} - 1)
 +
    (-\frac{1}{2} - 2)
 +
    \cdots
 +
    (-\frac{1}{2} - (100 - 1))
 +
  }{\cancel{(100)(99)\cdots(1)}}
 +
}{
 +
  \dfrac{
 +
    (\frac{1}{2})
 +
    (\frac{1}{2} - 1)
 +
    (\frac{1}{2} - 2)
 +
    \cdots
 +
    (\frac{1}{2} - (100 - 1))
 +
  }{\cancel{(100)(99)\cdots(1)}}
 +
} \\
 +
&= \frac{
 +
    (-\frac{1}{2})
 +
    (-\frac{1}{2} - 1)
 +
    (-\frac{1}{2} - 2)
 +
    \cdots
 +
    (-\frac{1}{2} - 99)
 +
}{
 +
    (\frac{1}{2})
 +
    (\frac{1}{2} - 1)
 +
    (\frac{1}{2} - 2)
 +
    \cdots
 +
    (\frac{1}{2} - 99)
 +
}
 +
\end{align*}</cmath>
 +
 +
Now, note that <math>-\frac{1}{2}-1=\frac{1}{2}-2</math>, <math>-\frac{1}{2}-2=\frac{1}{2}-3</math>, etc.; in essence, <math>-\frac{1}{2}-n=\frac{1}{2}-(n+1)</math>. We can then simplify the numerator and cancel like terms.
 +
 +
<cmath>\begin{align*}
 +
\frac{
 +
    (-\frac{1}{2})
 +
    (-\frac{1}{2} - 1)
 +
    (-\frac{1}{2} - 2)
 +
    \cdots
 +
    (-\frac{1}{2} - 99)
 +
}{
 +
    (\frac{1}{2})
 +
    (\frac{1}{2} - 1)
 +
    (\frac{1}{2} - 2)
 +
    \cdots
 +
    (\frac{1}{2} - 99)
 +
}
 +
&= \frac{
 +
    \cancel{(\frac{1}{2} - 1)}
 +
    \cancel{(\frac{1}{2} - 2)}
 +
    \cancel{(\frac{1}{2} - 3)}
 +
    \cdots
 +
    (\frac{1}{2} - 100)
 +
}{
 +
    (\frac{1}{2})
 +
    \cancel{(\frac{1}{2} - 1)}
 +
    \cancel{(\frac{1}{2} - 2)}
 +
    \cdots
 +
    \cancel{(\frac{1}{2} - 99)}
 +
} \\
 +
&= \frac{\frac{1}{2}-100}{\frac{1}{2}} \\
 +
&= \frac{-\frac{199}{2}}{\frac{1}{2}} \\
 +
&= \boxed{\textbf{(A) } -199.}
 +
\end{align*}</cmath>
  
 
== See also ==
 
== See also ==

Latest revision as of 17:17, 26 February 2018

Problem

For any real number a and positive integer k, define

${a \choose k} = \frac{a(a-1)(a-2)\cdots(a-(k-1))}{k(k-1)(k-2)\cdots(2)(1)}$

What is

${-\frac{1}{2} \choose 100} \div {\frac{1}{2} \choose 100}$?

$\textbf{(A)}\ -199\qquad \textbf{(B)}\ -197\qquad \textbf{(C)}\ -1\qquad \textbf{(D)}\ 197\qquad \textbf{(E)}\ 199$

Solution

We expand both the numerator and the denominator.

\begin{align*} \binom{-\frac{1}{2}}{100}\div\binom{\frac{1}{2}}{100} &= \frac{ \dfrac{ (-\frac{1}{2}) (-\frac{1}{2} - 1) (-\frac{1}{2} - 2) \cdots (-\frac{1}{2} - (100 - 1)) }{\cancel{(100)(99)\cdots(1)}} }{ \dfrac{ (\frac{1}{2}) (\frac{1}{2} - 1) (\frac{1}{2} - 2) \cdots (\frac{1}{2} - (100 - 1)) }{\cancel{(100)(99)\cdots(1)}} } \\ &= \frac{ (-\frac{1}{2}) (-\frac{1}{2} - 1) (-\frac{1}{2} - 2) \cdots (-\frac{1}{2} - 99) }{ (\frac{1}{2}) (\frac{1}{2} - 1) (\frac{1}{2} - 2) \cdots (\frac{1}{2} - 99) } \end{align*}

Now, note that $-\frac{1}{2}-1=\frac{1}{2}-2$, $-\frac{1}{2}-2=\frac{1}{2}-3$, etc.; in essence, $-\frac{1}{2}-n=\frac{1}{2}-(n+1)$. We can then simplify the numerator and cancel like terms.

\begin{align*} \frac{ (-\frac{1}{2}) (-\frac{1}{2} - 1) (-\frac{1}{2} - 2) \cdots (-\frac{1}{2} - 99) }{ (\frac{1}{2}) (\frac{1}{2} - 1) (\frac{1}{2} - 2) \cdots (\frac{1}{2} - 99) } &= \frac{ \cancel{(\frac{1}{2} - 1)} \cancel{(\frac{1}{2} - 2)} \cancel{(\frac{1}{2} - 3)} \cdots (\frac{1}{2} - 100) }{ (\frac{1}{2}) \cancel{(\frac{1}{2} - 1)} \cancel{(\frac{1}{2} - 2)} \cdots \cancel{(\frac{1}{2} - 99)} } \\ &= \frac{\frac{1}{2}-100}{\frac{1}{2}} \\ &= \frac{-\frac{199}{2}}{\frac{1}{2}} \\ &= \boxed{\textbf{(A) } -199.} \end{align*}

See also

1988 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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

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