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Algebra Difficulty 4.1 AIME Find the answer South Africa

What is the value of (2+4+6+...+198+200)(1+3+5+...+197+199)(2+4+6+...+198+200) - (1+3+5+...+197+199)?

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Rearranging, the value is (21)+(43)+(65)++(200199)(2-1) + (4-3) + (6-5) + \dots + (200-199) which has 100100 brackets and therefore totals 100×1=100100 \times 1 = 100.

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