\(\left(20^2+18^2+16^2+......+4^2+2^2\right)-\left(19^2+17^2+.....+3^2+1^2\right)\)

\(=20^2-19^2+18^2-17^2+......+2^2-1^2\)

\(=\left(20-19\right)\left(20+19\right)+\left(18-17\right)\left(18+17\right)+.......+\left(2-1\right)\left(2+1\right)\)

\(=39+35+....+7+3\)

\(=\left(39+3\right)\left[\left(39-3\right):4+1\right]:2=210\)

Ta có : A = 1.2 + 2.3 + 3.4 + ...... + 19.20

=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 19.20.21

=> 3A = 19.20.21

=> A = 19.20.21 / 3

=> A = 2660

1660 bạn nha

CHÚC BẠN HỌC GIỎI

\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{18.19}+\frac{2}{19.20}\)

\(=2\cdot\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)

\(=2\cdot\left(1-\frac{1}{20}\right)\)

\(=2\cdot\frac{19}{20}=\frac{19}{10}\)

44100 đó bạn

Nhớ k nhé

Ta có công thức :

Với mọi n thuộc N thì :

\(1^2+2^2+3^2+.......+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)

Áp dụng vào bài toán ta được :

\(A=1^2+2^2+3^2+....+20^2=\frac{20\left(20+1\right)\left(2.20+1\right)}{6}=2870\)

\(A=\dfrac{1}{20}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{19\cdot20}\right)\)

\(=\dfrac{1}{20}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)

=1/20-19/20=-18/20=-9/10