\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+......+\frac{1}{48.49.50}\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{48.49}-\frac{1}{49.50}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{49.50}\right)\)

\(=\frac{1}{2}.\frac{612}{1225}=\frac{612}{2450}=\frac{306}{1225}\)

Do not ask why hay quá!

Đặt \(T=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\)

Ta xét:

\(\frac{1}{1.2}-\frac{1}{2.3}=\frac{1}{1.2.3}\);\(\frac{1}{2.3}-\frac{1}{3.4}=\frac{1}{2.3.4}\);. . . ; \(\frac{1}{48.49}-\frac{1}{49.50}=\frac{1}{48.49.50}\)

 Rút ra dạng tổng quát,ta có: (mình nói thêm nhé)

\(\frac{1}{n\left(n+1\right)}-\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)\left(n+2\right)}\)

\(\Rightarrow2T=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\)

Ta nhận thấy: \(-\frac{1}{2.3}+\frac{1}{2.3}=0\);\(-\frac{1}{3.4}+\frac{1}{3.4}=0\);.....

\(\Rightarrow2T=\frac{1}{1.2}-\frac{1}{49.50}=\frac{612}{1225}\)

\(\Rightarrow T=\frac{612}{\frac{1225}{2}}=\frac{306}{1225}\)

Vậy .. . . 

ta co:1/1*2*3=(1/1*2-1/2*3):2
1/2*3*4=(1/1*2-1/2*3):2
...
cu nhu the cho den:
1/98*99*100=(1/98*99-1/99*100):2
suy ra : 1/1*2*3+1/2*3*4+1/3*4*5+...+1/98*99*100
=(1/1*2-1/2*3):2+(1/2*3-1/3*4):2+...+(1/98*99-1/99*100):2
=(1/1*2-1/2*3+1/2*3-1/3*4+...+1/98*99-1/99*100):2
=(1/1*2-1/99*100):2
=(1/2-1/9900)
=(4950/9000-1/9000):2
=4949/9000:2
=4949/18000
học tốt

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

Đặt A = 1 x 2 x 3 + 2 x 3 x 4 + 3 x 4 x 5 +....+ 98 x 99 x 100

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

4A = 1 x 2 x 3 x ( 4 - 0 ) + 2 x 3 x 4 x ( 5 - 1 ) + 4 x 5 x 6 x ( 7 - 3 ) +....+ 98 x 99 x 100 x ( 101 - 97 )

4A = 1 x 2 x 3 x 4 + 2 x 3 x 4 x 5 - 1 x 2 x 3 x 4 + 4 x 5 x 6 x 7 - 3 x 4 x 5 x 6 + .... + 98 x 99 x 100 x 101 - 98 x 99 x 100 x 97

A = 98 x 99 x 100 x 97 / 4

A = 98 x 99 x 25 x 97

4a=1.2.3.4+2.3.4(5-1)+3.4.5(6-2)+........+98.99.100(101-97)

4a=1.2.3.4+2.3.4.5-1.2.3.4+......+98.99.100.101-97.98.99.100

4a=98.99.100.101

a=(98.99.100.101):4=24497550