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Algebra Difficulty 4.7 AIME Find the answer Estonia

Find the value of
1⋅33⋅5+2⋅45⋅7+3⋅57⋅9+4⋅69⋅11+⋯+1009⋅10112019⋅2021 \frac{1 \cdot 3}{3 \cdot 5} + \frac{2 \cdot 4}{5 \cdot 7} + \frac{3 \cdot 5}{7 \cdot 9} + \frac{4 \cdot 6}{9 \cdot 11} + \dots + \frac{1009 \cdot 1011}{2019 \cdot 2021}

A number or a short expression. Spacing and $ signs are ignored.

Solution

Answer: 505⋅10092021=5095452021=252+2532021\frac{505 \cdot 1009}{2021} = \frac{509545}{2021} = 252 + \frac{253}{2021}.

The sum consists of 10091009 terms, where the ii-th term is of the form i(i+2)(2i+1)(2i+3)\frac{i(i+2)}{(2i+1)(2i+3)}. Let ss be the desired sum. Notice that
i(i+2)(2i+1)(2i+3)=14−34⋅1(2i+1)(2i+3)=14−38⋅(12i+1−12i+3) \frac{i(i+2)}{(2i+1)(2i+3)} = \frac{1}{4} - \frac{3}{4} \cdot \frac{1}{(2i+1)(2i+3)} = \frac{1}{4} - \frac{3}{8} \cdot \left( \frac{1}{2i+1} - \frac{1}{2i+3} \right)
Therefore
s=1009⋅14−38⋅((13−15)+(15−17)+⋯+(12019−12021))=10094−38⋅(13−12021)=10094−38⋅2021−33⋅2021=10094−20188⋅2021=10094−10094⋅2021=10094⋅(1−12021)=10094⋅20202021=505⋅10092021=5095452021 \begin{aligned} s &= 1009 \cdot \frac{1}{4} - \frac{3}{8} \cdot \left( \left( \frac{1}{3} - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{7} \right) + \dots + \left( \frac{1}{2019} - \frac{1}{2021} \right) \right) \\ &= \frac{1009}{4} - \frac{3}{8} \cdot \left( \frac{1}{3} - \frac{1}{2021} \right) = \frac{1009}{4} - \frac{3}{8} \cdot \frac{2021-3}{3 \cdot 2021} = \frac{1009}{4} - \frac{2018}{8 \cdot 2021} \\ &= \frac{1009}{4} - \frac{1009}{4 \cdot 2021} = \frac{1009}{4} \cdot \left( 1 - \frac{1}{2021} \right) = \frac{1009}{4} \cdot \frac{2020}{2021} = \frac{505 \cdot 1009}{2021} = \frac{509545}{2021} \end{aligned}

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.