Đặt \(A=1.4+2.5+3.6+...+100.103\)

\(=1\left(2.2\right)+2\left(3+2\right)+3\left(4+2\right)+...+100\left(101+2\right)\)

\(=1.2+2.3+3.4+...+100.101+\left(1.2+2.2+3.2+...+100.2\right)\)

\(=1.2+2.3+3.4+...+100.101+2\left(1+2+3+...+100\right)\)

\(=1.2+2.3+3.4+...+100.101+2.100\left(100+1\right):2\)

\(=1.2+2.3+3.4+...+100.101+10100\)

Đặt \(B=1.2+2.3+3.4+...+100.101\)

\(\Rightarrow3B=1.2.3+2.3.3+3.4.3+100.101.3\)

\(\Rightarrow3B=1.2.3+2.3\left(4-1\right)+3.4\left(5-2\right)+...+100.101\left(102-99\right)\)

\(\Rightarrow3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+100.101.102-99.100.101\)

\(\Rightarrow3B=100.101.102\)

\(\Rightarrow B=343400\)

Khi đó \(A=343400=10100=333300\)

Đặt A = 1.4 + 2.5 + 3.6 + 4.7 + ... + 100.103

3A = 3.(1.2 + 2.3 + 3.4 + ... + 100.101] + 3.(2 + 4 + 6 + ... + 200)

     = 1.2.3 + 2.3.3 + 3.4.3 + ... + 100.101.3 + 3.(2 + 4 + 6 + ... + 200)

\(\Rightarrow\) A  =  100.101.105:3 = 353500

Ta có: \(A=1\cdot99+2\cdot98+3\cdot97+\cdots+98\cdot2+99\cdot1\)

\(=2\left(1\cdot99+2\cdot98+\cdots+49\cdot51\right)+50\cdot50\)

\(=2\left\lbrack1\left(100-1\right)+2\left(100-2\right)+\cdots+49\left(100-49\right)\right\rbrack+2500\)

\(=2\cdot\left\lbrack100\left(1+2+\cdots+49\right)-\left(1^2+2^2+\cdots+49^2\right)\right\rbrack+2500\)

\(=2\cdot\left\lbrack100\cdot\frac{49\cdot50}{2}-\frac{49\cdot\left(49+1\right)\left(2\cdot49+1\right)}{6}\right\rbrack+2500\)

\(=2\left\lbrack50\cdot49\cdot50-\frac{49\cdot50\cdot99}{6}\right\rbrack+2500\)

\(=2\cdot\left\lbrack49\cdot50\cdot50-49\cdot25\cdot33\right\rbrack+2500\)

\(=2\cdot49\cdot25\cdot\left(2\cdot50-33\right)+2500\)

\(=49\cdot50\cdot67+2500=166650\)

Ta có: \(B=1\cdot2\cdot3+2\cdot3\cdot4+\ldots+17\cdot18\cdot19\)

\(=2\left(2-1\right)\left(2+1\right)+3\left(3-1\right)\left(3+1\right)+\cdots+18\left(18-1\right)\left(18+1\right)\)

\(=2\cdot\left(2^2-1\right)+3\left(3^2-1\right)+\cdots+18\left(18^2-1\right)\)

\(=\left(2^3+3^3+\cdots+18^3\right)-\left(2+3+\cdots+18\right)\)

\(=\left(1^3+2^3+\cdots+18^3\right)-\left(1+2+3+\cdots+18\right)\)

\(=\left(1+2+\cdots+18\right)^2-\left(1+2+\cdots+18\right)\)

\(=\left(18\cdot\frac{19}{2}\right)^2-18\cdot\frac{19}{2}=\left(9\cdot19\right)^2-9\cdot19=29070\)

Ta có: \(C=1\cdot4+2\cdot5+\cdots+100\cdot103\)

\(=1\left(1+3\right)+2\left(2+3\right)+\cdots+100\cdot\left(100+3\right)\)

\(=\left(1^2+2^2+\cdots+100^2\right)+3\left(1+2+\cdots+100\right)\)

\(=\frac{100\left(100+1\right)\left(2\cdot100+1\right)}{6}+\frac{3\cdot100\cdot101}{2}\)

\(=\frac{100\cdot101\cdot201}{6}+\frac{3\cdot100\cdot101}{2}=50\cdot101\cdot67+3\cdot50\cdot101\)

\(=50\cdot101\cdot70=3500\cdot101=353500\)

Ta có: \(D=1\cdot3+2\cdot4+3\cdot5+\cdots+97\cdot99+98\cdot100\)

\(=1\left(1+2\right)+2\left(2+2\right)+3\left(3+2\right)+\cdots+97\cdot\left(97+2\right)+98\cdot\left(98+2\right)\)

\(=\left(1^2+2^2+\cdots+98^2\right)+2\cdot\left(1+2+3+\cdots+98\right)\)

\(=\frac{98\cdot\left(98+1\right)\left(2\cdot98+1\right)}{6}+2\cdot\frac{98\cdot99}{2}\)

\(=\frac{98\cdot99\cdot197}{6}+98\cdot99=49\cdot33\cdot197+98\cdot99=49\cdot33\left(197+2\cdot3\right)\)

\(=49\cdot33\cdot203=328251\)

D =1.4+2.5+3.6+.......+99.102

D = 1. (2+2) +2.(2+3) +3.(2+4)+...+99.(100+2)

D = 1.2+1.2+2.2+2.3+2.3+3.4+...+2.99+99.100

D = (1.2+2.3+3.4+...+99.100) +2.(1+2+3+4+...+99)

*Gọi A= 1.2+2.3+3.4+...+99.100

3A = 3.(1.2+2.3+3.4+...+99.100)

3A = 1.2.3+2.3.3+...+99.100.3

3A = 1.2.3 +2 .3.(4-1)+...+99.100.(101-98)

3A = 1.2.3+2.3.4-1.2.3+...+ 99.100.101-98.99.100

3A = 99.100.101

3A = 3.33.100.101

A   =  33.100.101

A   = 333300

* Gọi B = 2. (1+2+3+4+...+99) 

                  \__có 99 số hạng ___/          

 B=  2.[(1+99).99:2]

 B = 2 .4950

 B = 9900

A+B = 333300+9900 =343200

Vậy D =343200