Difference between revisions of "1998 AIME Problems/Problem 5"
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== Problem == | == Problem == | ||
| + | Given that <math>A_k = \frac {k(k - 1)}2\cos\frac {k(k - 1)\pi}2,</math> find <math>|A_{19} + A_{20} + \cdots + A_{98}|.</math> | ||
== Solution == | == Solution == | ||
| + | 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. | ||
| + | |||
| + | 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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| + | 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. | ||
| + | |||
| + | 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>. | ||
== See also == | == See also == | ||
| − | + | {{AIME box|year=1998|num-b=4|num-a=6}} | |
| + | |||
| + | [[Category:Intermediate Trigonometry Problems]] | ||
Revision as of 17:23, 7 September 2007
Problem
Given that 
find 
Solution
Though the problem may appear to be quite daunting, it is actually not that difficult. 
always evaluates to an integer (triangular number), and the cosine of 
where 
is 1 if 
is even and -1 if is odd. will be even if 
or 
, and odd otherwise.
So our sum looks something like:
If we group the terms in pairs, we see that we need a formula for 
. So the first two fractions add up to 19, the next two to 
, and so forth.
If we pair the terms again now, each pair adds up to 
. There are 
such pairs, so our answer is 
.
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 | ||
