Tính nhanh a)(-1)×(-1)²×(-1)³×……×(-1)²⁰¹⁹ b)A=1×2+2×3+3×4+…..+1999×2000

$\begin{array}{l}
a)\,\,\left( { – 1} \right).{\left( { – 1} \right)^2}.{\left( { – 1} \right)^3}……{\left( { – 1} \right)^{2019}}\\
 = \left[ {\left( { – 1} \right){{\left( { – 1} \right)}^3}{{\left( { – 1} \right)}^5}……{{\left( { – 1} \right)}^{2019}}} \right]\left[ {{{\left( { – 1} \right)}^2}{{\left( { – 1} \right)}^4}…..{{\left( { – 1} \right)}^{2018}}} \right]\\
 = \underbrace {\left( { – 1} \right)\left( { – 1} \right)…..\left( { – 1} \right)}_{1010\,\,\,so\,\,\,\left( { – 1} \right)}.\underbrace {1.1.1……1}_{1009\,\,\,\,so\,\,1}\\
 = {\left( { – 1} \right)^{1010}}{.1^{1009}}\\
 = 1.1 = 1.\\
b)\,\,\,A = 1.2 + 2.3 + 3.4 + ….. + 1999.2000\\
 = 1.\left( {1 + 1} \right) + 2.\left( {1 + 2} \right) + 3.\left( {1 + 3} \right) + …. + 1999.\left( {1 + 1999} \right)\\
 = 1 + {1^2} + 2.1 + {2^2} + 3.1 + {3^2} + ….. + 1999 + {1999^2}\\
 = \left( {1 + 2 + 3 + ….. + 1999} \right) + \left( {{1^2} + {2^2} + {3^2} + ….. + {{1999}^2}} \right)\\
 = \frac{{1999\left( {1999 + 1} \right)}}{2} + \frac{{1999\left( {1999 + 1} \right)\left( {2.1999 + 1} \right)}}{6}\\
 = 1999.1000 + 1999.1333.1000\\
 = 1999.1000\left( {1 + 1333} \right)\\
 = 1334.1999.1000.
\end{array}$