D=1.2+2.3+3.4+...+19.20

=>3D=1.2.3+2.3.3+3.4.3+...+19.20

=1.2.3+2.3(4-1)+3.4(5-2)+...+19.20(21-18)

=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+19.20.21-18.19.20

=19.20.21=7980

=>D=7980:3=2660

Vậy D=2660

D=1.2+2.3+3.4+..+19.20

=>3D=1.2.3+2.3.3+3.4.3+...+19.20.3

=1.2.3+2.3(4-1)+3.4(5-2)+...+19.20(21-18)

=>1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+19.20.21-18.19.20

=19.20.21=7980

=>D=2660

vậy D=2660

A=1.2+2.3+3.4+.....+100.101

3A=1.2.3+2.3.3+3.4.3+..+100.101.3

3A=1.2.3+2.3.(4-1)+3.4.(5-2)......+100.101.(102-99)

3A=1.2.3+2.3.4-2.3.1+3.4.5-3.4.2+....+99.100.101-100.101.102

3A=100.101.102

3A=\(\frac{100.101.101}{3}\)

Ta có : A = 1.2 + 2.3 + 3.4 + ..... + 100.101

=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 100.101.102

=> 3A = 100.101.102

=> A = 100.101.102/3

=> A = 343400

Ta có công thức tổng quát : \(f\left(x\right)=1.2+2.3+3.4+...+x\left(x+1\right)=\frac{x\left(x+1\right)\left(x+2\right)}{3}\)

Do vậy f(x) = 0 \(\Leftrightarrow\frac{x\left(x+1\right)\left(x+2\right)}{3}=0\Leftrightarrow x\left(x+1\right)\left(x+2\right)=0\)

Tới đây bạn tự làm! (chú ý rằng bạn chưa cho điều kiện của x)

A = 1.2 + 2.3 + 3.4 + ... + 100.101

3A = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) + ... + 100.101.(102-99)

3A = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 100.101.102 - 99.100.101

3A = (1.2.3 + 2.3.4 + 3.4.5 + ... + 100.101.102) - (0.1.2 + 1.2.3 + 2.3.4 + ... + 99.100.101)

3A = 100.101.102 - 0.1.2

3A = 100.101.102

A = 100.101.34

A = 343400