Difference between revisions of "1998 AIME Problems/Problem 5"
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== Solution == | == Solution == | ||
− | Though the problem may appear to be quite daunting, it is actually not that difficult. <math> | + | 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>\left|\sum_{i=19}^{98} A_i\right| = | + | <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> |
− | 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> | + | 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> | + | 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 == | ||
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[[Category:Intermediate Trigonometry Problems]] | [[Category:Intermediate Trigonometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 19:38, 4 July 2013
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
, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.