TÍNH hợp lý: 2020:[20:3}
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(-1) + 2 + (-3) + 4 + ... + (-2019) + 2020
= [ (-1) + 2 ] + [ (-3) + 4 ] + ... + [ (-2019) + 2020 ]
= 1 + 1 + ... + 1
= 1 . 1010
= 1010
(-1)+2+(-3)+4+...+(-2019)+2020
=(-1+2)+(-3+4)+...+(-2019+2020)
=1+1+...+1 (có 1010 số 1)
=1.1010
=1010
\(=\left(137-157\right)+\left(2020-40\right)=-20+1980=1960\)
..................................................................
b: Sửa đề: \(B=1\cdot3+2\cdot4+3\cdot5+4\cdot6+\cdots+2018\cdot2020\)
\(=1\left(1+2\right)+2\left(2+2\right)+\cdots+2018\left(2018+2\right)\)
\(=\left(1^2+2^2+\cdots+2018^2\right)+2\left(1+2+\cdots+2018\right)\)
\(=\frac{2018\left(2018+1\right)\left(2\cdot2018+1\right)}{6}+2\cdot\frac{2018\cdot2019}{2}\)
\(=\frac{2018\cdot2019\cdot4037}{6}+2018\cdot2019=1009\cdot673\cdot4037+2018\cdot2019\)
\(=1009\cdot673\left(4037+2\cdot3\right)=1009\cdot673\cdot4043\)
=2745427451
c: Ta có: \(C=\left(1+2+3+\cdots+2020\right)\left(20192019\cdot2020-2019\cdot20202020\right)+\left(15-12\right)^2\)
\(=\frac{2020\left(2020+1\right)}{2}\cdot2019\cdot2020\cdot\left(10001-10001\right)+3^2\)
=0+9
=9
b: Sửa đề: \(B=1\cdot3+2\cdot4+3\cdot5+4\cdot6+\cdots+2018\cdot2020\)
\(=1\left(1+2\right)+2\left(2+2\right)+\cdots+2018\left(2018+2\right)\)
\(=\left(1^2+2^2+\cdots+2018^2\right)+2\left(1+2+\cdots+2018\right)\)
\(=\frac{2018\left(2018+1\right)\left(2\cdot2018+1\right)}{6}+2\cdot\frac{2018\cdot2019}{2}\)
\(=\frac{2018\cdot2019\cdot4037}{6}+2018\cdot2019=1009\cdot673\cdot4037+2018\cdot2019\)
\(=1009\cdot673\left(4037+2\cdot3\right)=1009\cdot673\cdot4043\)
=2745427451
c: Ta có: \(C=\left(1+2+3+\cdots+2020\right)\left(20192019\cdot2020-2019\cdot20202020\right)+\left(15-12\right)^2\)
\(=\frac{2020\left(2020+1\right)}{2}\cdot2019\cdot2020\cdot\left(10001-10001\right)+3^2\)
=0+9
=9
\(\frac{1}{3}+\frac{2}{3^2}\frac{3}{3^3}+...+\frac{2019}{3^{2019}}\)