\(Đật.A=1+2\left(1+1\right)+3\left(2+1\right)+...+100\left(99+1\right)\\ =1+1\cdot2+2+2\cdot3+3+...+99\cdot100+100\\ =\left(1+2+...+100\right)+\left(1\cdot2+2\cdot3+...+99\cdot100\right)\)
Đặt \(M=1\cdot2+2\cdot3+...+99\cdot100\)
\(3M=1\cdot2\cdot3+2\cdot3\cdot3+...+99\cdot100\cdot3\\ 3M=1\cdot2\left(3-0\right)+2\cdot3\left(4-1\right)+...+99\cdot100\left(101-98\right)\\ 3M=1\cdot2\cdot3-1\cdot2\cdot3+2\cdot3\cdot4-...-98\cdot99\cdot100+99\cdot100\cdot101\\ 3M=99\cdot100\cdot101\\ M=333300\)
Vậy \(A=333300+\dfrac{\left(100+1\right)\cdot100}{2}=333300+5050=338050\)
= 1(2-1)+2(3-1)+3(4-1)+....+100(101-1)
=1.2-1.1+2.3-2.1+3.4-3.1+....+100.101-100
=(1.2+2.3+3.4+...+100.101)-(1+2+3+...+100)
=3(1.2+2.3+3.4+...+100.101)/3 -(1+2+3+..+100)
=[1.2.3+2.3.(4-1)+3.4.(5-2)+...+100.101(102-99)]/3 -(1+2+3+..+100)
=(1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+100.101.102-99.100.101)/3-5050
=\(\dfrac{100.101.102}{3}\)-5050=343400-5050=338350
mik làm vậy nếu sai đừng trách mik 
