\(B=\frac{\frac{2016}{1000}+\frac{2016}{999}+\frac{2016}{998}+...+\frac{2016}{501}}{-\frac{1}{1\cdot2}-\frac{1}{3\cdot4}-\frac{1}{5\cdot6}-...-\frac{1}{999\cdot1000}}\)

\(B=\frac{2016\left(\frac{1}{1000}+\frac{1}{999}+\frac{1}{998}+...+\frac{1}{501}\right)}{-\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{999\cdot1000}\right)}\)

\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{999}-\frac{1}{1000}\right)}\)

\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left[\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\right)\right]}\)

\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1000}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\right)\right]}\)

\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1000}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{500}\right)}\)

\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}\)

\(B=\frac{2016}{-1}=-2016\)

cảm ơn bạn Phương Uyên

A=(1-2)+(3-4)+...+(999-1000)

      có 1000 số hạng

A=(-1)+(*1)+...+(-1)

     có 500 số hạng

A=-1*500

A=-500

P(x) = x²⁰¹⁹ - 1000x²⁰¹⁸ + 1000x²⁰¹⁷ - 1000x²⁰¹⁶ + ... + 1000x - 1

Với x = 999 => 1000 = x + 1

=> P(999) = x²⁰¹⁹ - ( x + 1 )x²⁰¹⁸ + ( x + 1 )x²⁰¹⁷ - ( x + 1 )x²⁰¹⁶ + ... + ( x + 1 )x - 1

= x²⁰¹⁹ - x²⁰¹⁹ - x²⁰¹⁸ + x²⁰¹⁸ + x²⁰¹⁷ - x²⁰¹⁷ - x²⁰¹⁶ + ... + x² + x - 1

= x - 1 = 999 - 1 = 998

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