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
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 | |
| 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 | ||
