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

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

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

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

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

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

E = 1.3 + 2.4 + 3.5 +...+ 97.99 + 98.100

A = 1.3 + 3.5 + 5.7 + ...+ 97.99

B = 2.4 + 4.6 + 6.8 + ... + 98.100

A = 1.3 + 3.5 + 5.7 + ... + 97.99

6A = 1.3.6 + 3.5.6 + 5.7.6 +...+ 97.99.6

1.3.6 = 1.3.(5+ 1) = 1.3.5 + 1.3.1

3.5.6 = 3.5(7 - 1) = 3.5.7 - 1.3.5

5.7.6 = 5.7.(9 - 3) = 5.7.9 - 3.5.7

7.9.6 = 7.9.(11 - 5) = 7.9.11 - 5.7.9

..........................................................................

97.99.6 = 97.99.(101 - 95) = 97.99.101 - 95.97.99

Cộng vế với vế ta có:

6A = 1.3.1 + 97.99.101

6A = 3 + 969903

6A = 969906

A = 969906 : 6

A = 161651

B = 2.4 + 4.6 + 6.8 + ... + 98.100

6B = 2.4.6 + 4.6.6 + 6.8.6 + ... + 98.100.6

2.4.6 = 2.4.6

4.6.6 = 4.6.(8 - 2) = 4.6.8 - 2.4.6

6.8.6 = 6.8.(10 - 4) = 6.8.10 - 4.6.8

8.10.6 = 8.10.(12 - 6) = 8.10.12 - 6.8.10

...............................................................................

98.100.6 = 98.100.(102 - 96) = 98.100.102 - 96.98.100

6B = 98.100.102

B = 98.100.102 : 6

B = 166600

E = A + B

E = 161651 + 166600

E = 328251

A = 1.4 + 2.5 + 3.6 + ... + 99.102

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

A = (1.2 + 2.3 + 3.4 + ... + 99.100) + (1.2 + 2.2 + 3.2 + ... + 99.2)

Đặt B = 1.2 + 2.3 + 3.4 + ... + 99.100

3B = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) + ... + 99.100.(101-98)

3B = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100

3B = 99.100.101

B = 33.100.101 = 333300

A = 333300 + 2.(1 + 2 + 3 + ... + 99)

A = 333300 + 2.(1 + 99).99:2

A = 333300 + 100.99

A = 333300 + 9900

A = 343200

 A = 1.4 + 2.5 + 3.6 +...+ 99.102

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

A = 1.2+1.2+2.3+2.2+3.4+3.2+.+99.100+99.2

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

Ta thấy: 1.4 = 1.(1 + 3)

2.5 = 2.(2 + 3)

3.6 = 3.(3 + 3)

4.7 = 4.(4 + 3)

…….

n(n + 3) = n(n + 1) + 2n

Vậy C = 1.2 + 2.1 + 2.3 + 2.2 + 3.4 + 2.3 + … + n(n + 1) +2n

C = 1.2 + 2 +2.3 + 4 + 3.4 + 6 + … + n(n + 1) + 2n

C = [1.2 +2.3 +3.4 + … + n(n + 1)] + (2 + 4 + 6 + … + 2n)

⇒ 3C = 3.[1.2 +2.3 +3.4 + … + n(n + 1)] + 3.(2 + 4 + 6 + … + 2n) 

3C = 1.2.3 + 2.3.3 + 3.4.3 + … + n(n + 1).3 + 3.(2 + 4 + 6 + … + 2n)

3C = n(n + 1)(n + 2) + 

⇒ C =  +  = 

Ta thấy: 1.4 = 1.(1 + 3)

2.5 = 2.(2 + 3)

3.6 = 3.(3 + 3)

4.7 = 4.(4 + 3)

…….

n(n + 3) = n(n + 1) + 2n

Vậy C = 1.2 + 2.1 + 2.3 + 2.2 + 3.4 + 2.3 + … + n(n + 1) +2n

C = 1.2 + 2 +2.3 + 4 + 3.4 + 6 + … + n(n + 1) + 2n

C = [1.2 +2.3 +3.4 + … + n(n + 1)] + (2 + 4 + 6 + … + 2n)

⇒ 3C = 3.[1.2 +2.3 +3.4 + … + n(n + 1)] + 3.(2 + 4 + 6 + … + 2n) 

3C = 1.2.3 + 2.3.3 + 3.4.3 + … + n(n + 1).3 + 3.(2 + 4 + 6 + … + 2n)

3C = n(n + 1)(n + 2) + 

⇒ C =  +  = 

\(A=1\cdot4+2\cdot5+3\cdot6+...+n\left(n+3\right)\)

\(=1\left(1+3\right)+2\left(2+3\right)+3\left(3+3\right)+...+n\left(n+3\right)\)

\(=\left(1^2+2^2+...+n^2\right)+3\left(1+2+3+...+n\right)\)

\(=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}+3\cdot\dfrac{n\left(n+1\right)}{2}\)

\(=\dfrac{n\left(n+1\right)\left(2n+1\right)+9n\left(n+1\right)}{6}\)

\(=\dfrac{n\left(n+1\right)\left(2n+1+9\right)}{6}\)

\(=\dfrac{n\left(n+1\right)\left(2n+10\right)}{6}=\dfrac{n\left(n+1\right)\left(n+5\right)}{3}\)