Difference between revisions of "1999 AHSME Problems/Problem 1"
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== Solution == | == Solution == | ||
+ | === Solution 1 === | ||
If we group consecutive terms together, we get <math>(-1) + (-1) + \cdots + 99</math>, and since there are 49 pairs of terms the answer is <math>-49 + 99 = 50 \Rightarrow \mathrm{(E)}</math>. | If we group consecutive terms together, we get <math>(-1) + (-1) + \cdots + 99</math>, and since there are 49 pairs of terms the answer is <math>-49 + 99 = 50 \Rightarrow \mathrm{(E)}</math>. | ||
+ | |||
+ | === Solution 2 === | ||
+ | Let <math>1 - 2 + 3 -4 + \cdots - 98 + 99 = S</math>. | ||
+ | |||
+ | Therefore, <math>S=99-98+97-\cdots -4+3-2+1</math> | ||
+ | |||
+ | We add: | ||
+ | |||
+ | <math>2S=100-100+100-100\cdots +100=100</math> | ||
+ | |||
+ | <math>S=50\Rightarrow \mathrm{(E)}</math> | ||
== See also == | == See also == |
Revision as of 08:15, 20 March 2008
Contents
[hide]Problem
Solution
Solution 1
If we group consecutive terms together, we get , and since there are 49 pairs of terms the answer is .
Solution 2
Let .
Therefore,
We add:
See also
1999 AHSME (Problems • Answer Key • Resources) | ||
Preceded by First question |
Followed by Problem 2 | |
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All AHSME Problems and Solutions |