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Difference between revisions of "1998 AIME Problems/Problem 5"

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== Solution ==
 
== Solution ==
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Though the problem may appear to be quite daunting, it is actually not that difficult. <math>\displaystyle \frac {k(k-1)}2</math> always evaluates to an integer ([[triangular number]]), and the [[cosine]] of <math>\displaystyle n\pi</math> where <math>n \in \mathbb{Z}</math> is 1 if <math>\displaystyle n</math> is even and -1 if <math>\displaystyle n</math> is odd. <math>\displaystyle \frac {k(k-1)}2</math> will be even if <math>\displaystyle 4|k</math> or <math>\displaystyle 4|k-1</math>, and odd otherwise.   
+
Though the problem may appear to be quite daunting, it is actually not that difficult. <math>\frac {k(k-1)}2</math> always evaluates to an integer ([[triangular number]]), and the [[cosine]] of <math>n\pi</math> where <math>n \in \mathbb{Z}</math> is 1 if <math>n</math> is even and -1 if <math>n</math> is odd. <math>\frac {k(k-1)}2</math> will be even if <math>4|k</math> or <math>4|k-1</math>, and odd otherwise.   
  
 
So our sum looks something like:
 
So our sum looks something like:
  
−
<math>|\sum_{i=19}^{98} A_i| = \displaystyle - \frac{19(18)}{2} + \frac{20(19)}{2} + \frac{21(20)}{2} - \frac{22(21)}{2} \cdots - \frac{98(97)}{2}</math>
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<math>\left|\sum_{i=19}^{98} A_i\right| =- \frac{19(18)}{2} + \frac{20(19)}{2} + \frac{21(20)}{2} - \frac{22(21)}{2} - \frac{23(22)}{2} + \frac{24(23)}{2} \cdots - \frac{98(97)}{2}</math>
  
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If we group the terms in pairs, we see that we need a formula for <math>\frac{n(n-1)}{2} + \frac{n+1(n)}{2} = \left(\frac n2\right)(n+1 - (n-1)) = n</math>. So the first two fractions add up to 19, the next two to <math>-21</math>, and so forth.
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If we group the terms in pairs, we see that we need a formula for <math>-\frac{(n)(n-1)}{2} + \frac{(n+1)(n)}{2} = \left(\frac n2\right)(n+1 - (n-1)) = n</math>. So the first two fractions add up to <math>19</math>, the next two to <math>-21</math>, and so forth.
  
−
If we pair the terms again now, each pair adds up to <math>-2</math>. There are <math>\frac{98-19+1}{2 \cdot 2} = 20</math> such pairs, so our answer is <math>|-2 \cdot 20| = 040</math>.
+
If we pair the terms again now, each pair adds up to <math>-2</math>. There are <math>\frac{98-19+1}{2 \cdot 2} = 20</math> such pairs, so our answer is <math>|-2 \cdot 20| = \boxed{040}</math>.
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Trigonometry Problems]]
 
[[Category:Intermediate Trigonometry Problems]]
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{{MAA Notice}}

Revision as of 01:20, 16 September 2017

Problem

Given that $A_k = \frac {k(k - 1)}2\cos\frac {k(k - 1)\pi}2,$ find $|A_{19} + A_{20} + \cdots + A_{98}|.$

Solution

Though the problem may appear to be quite daunting, it is actually not that difficult. $\frac {k(k-1)}2$ always evaluates to an integer (triangular number), and the cosine of $n\pi$ where $n \in \mathbb{Z}$ is 1 if $n$ is even and -1 if $n$ is odd. $\frac {k(k-1)}2$ will be even if $4|k$ or $4|k-1$, and odd otherwise.

So our sum looks something like:

$\left|\sum_{i=19}^{98} A_i\right| =- \frac{19(18)}{2} + \frac{20(19)}{2} + \frac{21(20)}{2} - \frac{22(21)}{2} - \frac{23(22)}{2} + \frac{24(23)}{2} \cdots - \frac{98(97)}{2}$

If we group the terms in pairs, we see that we need a formula for $-\frac{(n)(n-1)}{2} + \frac{(n+1)(n)}{2} = \left(\frac n2\right)(n+1 - (n-1)) = n$. So the first two fractions add up to $19$, the next two to $-21$, and so forth.

If we pair the terms again now, each pair adds up to $-2$. There are $\frac{98-19+1}{2 \cdot 2} = 20$ such pairs, so our answer is $|-2 \cdot 20| = \boxed{040}$.

See also

1998 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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