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HMMT February

Algebra Difficulty 5.0 AIME Find the answer

Determine the value of 1⋅2−2⋅3+3⋅4−4⋅5+⋯+2001⋅20021 \cdot 2-2 \cdot 3+3 \cdot 4-4 \cdot 5+\cdots+2001 \cdot 2002

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

Solution

2004002. Rewrite the expression as 2+3⋅(4−2)+5⋅(6−4)+⋯+2001⋅(2002−2000)2+3 \cdot(4-2)+5 \cdot(6-4)+\cdots+2001 \cdot(2002-2000) =2+6+10+⋯+4002=2+6+10+\cdots+4002 This is an arithmetic progression with (4002−2)/4+1=1001(4002-2) / 4+1=1001 terms and average 2002, so its sum is 1001⋅2002=20040021001 \cdot 2002=2004002.

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