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A=1.2.3+2.3.4+...+8.9.10

4A=1.2.3.4+2.3.4.4+...+8.9.10.4

4A=1.2.3.(4-0)+2.3.4.(5-1)+...+8.9.10.(11-7)

4A=(1.2.3.4+2.3.4.5+...+8.9.10.11)-(0.1.2.3+1.2.3.4+...+7.8.9.10)

4A=8.9.10.11-0.1.2.3

4A=7920-0

4A=7920

A=7920:4

A=1980

Vậy A=1980

\(S=9\cdot10+10\cdot11+11\cdot12+...+99\cdot100\)

\(3S=9\cdot10\cdot3+10\cdot11\cdot3+11\cdot12\cdot3+...+99\cdot100\cdot3\)

\(3S=9\cdot10\cdot\left(11-8\right)+10\cdot11\cdot\left(12-9\right)+...+99\cdot100\cdot\left(101-98\right)\)

\(3S=9\cdot10\cdot11-8\cdot9\cdot10+10\cdot11\cdot12-9\cdot10\cdot11+...+99\cdot100\cdot101-98\cdot99\cdot100\)

\(3S=99\cdot100\cdot101\)

\(S=\frac{99\cdot100\cdot101}{3}=333300\)

A=1.2+2.(3.2)+2.(5.3)+...+2.(99.50)

A=2.(3.2+5.3+...+99.50)

c, 4C= (1.2.3+2.3.4+3.4.5+...+8.9.10) .4

==> 4C= [1.2.3.(4-0) + 2.3.4-(5-1) + 8.9.10.(11-7)

==>4C= 1.2.3.4 - 1.2.3.4+ 2.3.4.5-2.3.4.5 + 7.8.9.10- 7.8.9.10 + 8.9.10.11

==> 4C= 8.9.10.11=7920

==> C= 7920 :4=1980

a, Ta có: 3A= 1.2.3+2.3.3+3.4.3+...+99.100.3

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

               3A=(1.2.3 + 2.3.4+ 3.4.5+ 99.100.101) - (0.1.2 +1.2.3+ 2.3.4 + ... + 98.99.100)

               3A= 99.100.101 - 0.1.2

               3A= 999900 - 0

               3A= 999900

    ==> A= 999900 : 3

   ==> A= 333300

Đặt : \(A=10.11+11.12+...+98.99+99.100\)

\(\Rightarrow3A=10.11.3+11.12.3+...+98.99.3+99.100.3\)

\(\Rightarrow3A=10.11.\left(12-9\right)+11.12.\left(13-10\right)+...+\)\(98.99.\left(100-97\right)+99.100.\left(101-98\right)\)

SAI ĐỀ RỒI BẠN. SỬA 23=2.3

\(\frac{2.9.8+3.12.10+4.15.12+...+98.297.200}{2.3.4+3.4.5+4.5.6+...+98.99.100}\)

\(\frac{1.2.3.\left(2.3.4+3.4.5+4.5.6+...+98.99.100\right)}{\left(2.3.4+3.4.5+4.5.6+...+98.99.10\right)}\)

\(=6\)

VẬYa²=6²=36

\(A=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)

\(3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+99\cdot100\cdot\left(101-98\right)\)

\(3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+...+99\cdot100\cdot101-98\cdot99\cdot100\)

\(3A=99\cdot100\cdot101\Rightarrow A=\dfrac{99\cdot100\cdot101}{3}=333300\)

\(B=1^2+2^2+3^2+...+99^2+100^2\)

\(=\dfrac{100\cdot\left(100+1\right)\cdot\left(2\cdot100+1\right)}{6}\)

\(=\dfrac{2030100}{6}=338350\)

\(C=1\cdot2\cdot3+2\cdot3\cdot4+...+8\cdot9\cdot10\)

\(4C=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot\left(5-1\right)+...+8\cdot9\cdot10\cdot\left(11-7\right)\)

\(4C=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+...+8\cdot9\cdot10\cdot11-7\cdot8\cdot9\cdot10\)

\(4C=8\cdot9\cdot10\cdot11\Rightarrow C=\dfrac{8\cdot9\cdot10\cdot11}{4}=1980\)

@Hồng Phúc Nguyễn

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

3A = 1 . 2 . 3 + 2 . 3 . (4 - 1) + 3 . 4 . (5 - 2) + ... + 99 . 100 . (101 - 98)

3A = 1 . 2 . 3 + 2 . 3 . 4 - 2 . 3 . 1 + 3 . 4 . 5 - 3 . 4 . 2 + ... + 99 . 100 . 101 - 99 . 100 . 98

3A = 99 . 100 . 98

A = 33 . 100 . 98

A = 323400