đem mớ này nhồi vào đầu rồi đầy quá đứt mạch máu não , tử vong tại chỗ

ta có:
4s=1.2.3.(4-0)+2.3.4.(5-1)+3.4.5.(6-2)+.........+k(k+1)(k+2)((k+3)-(k-1))
4s=1.2.3.4-1.2.3.0+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+........+k(k+1)(k+2)(k+3)-(k-1)k(k+1)(k+2)
4s=k(k+1)(k+2)(k+3)
ta biết rằng tích 4 số tự nhiên liên tiếp khi cộng thêm 1 luôn là 1 số chính phương
=>4s+1 là 1 số chính phương

\(=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{18\cdot19}-\dfrac{1}{19\cdot20}\)

=1/2-1/380

=190/380-1/380

=189/380

Gọi biểu thức trên là S. Ta có :

\(S=\dfrac{1}{1\times2\times3}+\dfrac{1}{2\times3\times4}+\dfrac{1}{3\times4\times5}+...+\dfrac{1}{18\times19\times20}\)

\(=\dfrac{1}{2}\times\left(\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+...+\dfrac{2}{18\times19\times20}\right)\)

Trước tiên, ta áp dụng : \(\dfrac{2}{a\left(a+1\right)\left(a+2\right)}=\dfrac{1}{a\left(a+1\right)}-\dfrac{1}{\left(a+1\right)\left(a+2\right)}\)

Ta sẽ có : 

\(S=\dfrac{1}{2}\times\left(\dfrac{1}{1\times2}-\dfrac{1}{2\times3}+\dfrac{1}{2\times3}-\dfrac{1}{3\times4}+\dfrac{1}{3\times4}-\dfrac{1}{4\times5}+...+\dfrac{1}{18\times19}-\dfrac{1}{19\times20}\right)\)

\(=\dfrac{1}{2}\times\left(\dfrac{1}{1\times2}-\dfrac{1}{19\times20}\right)\)

\(=\dfrac{1}{2}\times\dfrac{1}{1\times2}-\dfrac{1}{2}\times\dfrac{1}{19\times20}\)

\(=\dfrac{1}{4}-\dfrac{1}{760}=\dfrac{189}{760}\)

`=1/2(1/1×2 - 1/2×3 + 1/2×3 - 1/3×4 + 1/3×4 - 1/4×5 + ... + 1/18×19 - 1/19×20)`
`=1/2(1/2 - 1/19×20)`
`=1/2×189/380 `
`=189/760`

Đặt \(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{98\cdot99\cdot100}\)

Ta có: \(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{98\cdot99\cdot100}\)

\(\Leftrightarrow2A=\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+\dfrac{2}{3\cdot4\cdot5}+...+\dfrac{2}{98\cdot99\cdot100}\)

\(\Leftrightarrow2A=-\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}-\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}-\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}-\dfrac{1}{4\cdot5}+...-\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)

\(\Leftrightarrow2A=-\dfrac{1}{2}+\dfrac{1}{99\cdot100}\)

\(\Leftrightarrow2A=\dfrac{-1}{2}+\dfrac{1}{9900}\)

\(\Leftrightarrow2A=\dfrac{-4950}{9900}+\dfrac{1}{9900}=\dfrac{-4949}{9900}\)

hay \(A=\dfrac{-4949}{19800}\)

\(C=1.2.3+2.3.4+........+48.49.50\)

\(\Rightarrow4C=1.2.3.4+2.3.4.4+........+48.49.50.4\)

\(=1.2.3.4+2.3.4.\left(5-1\right)+.........+48.49.50.\left(51-47\right)\)

\(=1.2.3.4+2.3.4.5-1.2.3.4+........+48.49.50.51-47.48.49.50\)

\(=48.49.50.51\)

\(\Rightarrow C=\frac{48.49.50.51}{4}=1499400\)

Ta có C = 1 x 2 x 3 + 2 x 3 x 4 + ... + 48 x 49 x 50

=>   4C  = 1 x 2 x 3 x 4 + 2 x 3 x 4 x 4 + ....  + 48 x 49 x 50 x 4

       4C   = 1 x 2 x 3 x 4 +  2 x 3 x 4 x (5 - 1)+ ... + 48 x 49 x 50 x (51 - 47)

       4C   = 1 x 2 x 3 x 4 + 2 x 3 x 4 x 5 - 1 x 2 x 3 x 4 + .... + 48 x 49 x 50 x 51 -  47 x 48 x 49 x 50 

       4C   = 48 x 49 x 50 x 51

       4C   = 5997600

         C   = 5997600 : 4

         C   = 1499400

Vậy C   = 1499400

C = 1x2x3+2x3x4+...+48x49x50

4C = 1x2x3x4+2x3x4x4+...+48x49x50

4C = 1x2x3x(4-0)+2x3x4x(5-1)+...+48x49x50x(51-47)

4C = 1x2x3x4-0x1x2x3+2x3x4x5-1x2x3x4+...+48x49x50x51-47x48x49x50

4C = 48x49x50x51

=> C = (48x49x50x51)/4

C = 12x49x50x51

A =1x2x3 + 2x3x4 +3x4x5+....+ 2010 x2011 x 2012

4A =1x2x3x4 + 2x3x4x4 +3x4x5x4+....+ 2010 x2011 x 2012x4

4A =1x2x3x4 + 2x3x4x(5+1) +3x4x5x(6-2)+....+ 2010 x2011 x 2012x(2013-2009)

4A =1x2x3x4 + 2x3x4x5-1x2x3x4+3x4x5x6-2x3x4x5+....+ 2010 x2011 x 2012x2013-2009x2010x2011x2012

4A = 2010 x2011 x 2012x2013

A = \(\frac{2010\times2011\times2012\times2013}{4}\)

đặt S=1.2.3+2.3.4+....+18.19.20

4S=1.2.3.4+2.3.4.(5-1)+.......+18.19.20.(21-17)

4S=1.2.3.4-1.2.3+2.3.4.5-1.2.3.4+......+18.19.20.21-17.18.19.20

4S=....tự làm nha

bonking copy đó

..đề đến 2010,2011,2012 à...

nhục thật ^^

haizz công nhận trên đời còn có mấy loại.....ok mk thì hay lắm mà

Để ngày mai mik hỏi cô giáo jup cho

\(A=1\cdot2\cdot3+2\cdot3\cdot4+...+7\cdot8\cdot9+8\cdot9\cdot10\)

\(4A=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot4+...+7\cdot8\cdot9\cdot4+8\cdot9\cdot10\cdot4\)

\(4A=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot\left(5-1\right)+...+7\cdot8\cdot9\cdot\left(10-6\right)+8\cdot9\cdot10\cdot\left(11-7\right)\)

\(4A=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+...+7\cdot8\cdot9\cdot10-6\cdot7\cdot8\cdot9+8\cdot9\cdot10\cdot11-7\cdot8\cdot9\cdot10\)

\(4A=8\cdot9\cdot10\cdot11\)

\(A=\frac{8\cdot9\cdot10\cdot11}{4}=1980\)

A=1·2·3+2·3·4+...+7·8·9+8·9·10

4A=1·2·3·4+2·3·4·4+...+7·8·9·4+8·9·10·4

4A=1·2·3·4+2·3·4·(5−1)+...+7·8·9·(10−6)+8·9·10·(11−7)

4A=1·2·3·4+2·3·4·5−1·2·3·4+...+7·8·9·10−6·7·8·9+8·9·10·11−7·8·9·10

4A=8·9·10·11