Not that large

Algebra Level 5

A = 1 1 × 2 + 1 3 × 4 + ⋯ + 1 2003 × 2004 + 1 2005 × 2006 B = 1 1004 × 2006 + 1 1005 × 2005 + ⋯ + 1 2006 × 1004 \begin{aligned} A&=&\frac{1}{1\times2}+\frac{1}{3\times4}+\cdots+\frac{1}{2003\times2004}+\frac{1}{2005\times2006} \\ B&=&\frac{1}{1004\times2006}+\frac{1}{1005\times2005}+\cdots+\frac{1}{2006\times1004} \end{aligned}

Find the value of A B {\frac{A}{B}} .


The answer is 1505.

Adam Phúc Nguyễn by Adam Phúc Nguyễn , 20, Vietnam

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1 solution

A = 1 − 1 2 + 1 3 − 1 4 + … + 1 2005 − 1 2006 A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{2005}-\frac{1}{2006} = ( 1 + 1 3 + … + 1 2005 ) − ( 1 2 + 1 4 + … + 1 2006 ) =\left(1+\frac{1}{3}+\ldots+\frac{1}{2005}\right)-\left(\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2006}\right) = ( 1 + 1 2 + 1 3 + … + 1 2015 ) − 2 ( 1 2 + 1 4 + … + 1 2006 ) =\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2015}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2006}\right) = 1 1004 + 1 1005 + … + 1 2006 =\frac{1}{1004}+\frac{1}{1005}+\ldots+\frac{1}{2006} B = 1 3010 ( 3010 1004 × 2006 + 3010 1005 × 2005 + … + 3010 2006 × 1004 ) B=\frac{1}{3010}\left(\frac{3010}{1004\times2006}+\frac{3010}{1005\times2005}+\ldots+\frac{3010}{2006\times1004} \right) = 2 3010 ( 1 1004 + 1 1005 + … + 1 2006 ) =\frac{2}{3010}\left(\frac{1}{1004}+\frac{1}{1005}+\ldots+\frac{1}{2006}\right) A B = 3010 2 = 1505 \frac{A}{B}=\frac{3010}{2}=\boxed{1505}

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