Question

\(\frac{1}{1x2x3}+\frac{1}{2x3x4}+\frac{1}{3x4x5}+...+\frac{1}{48x49x50}\)

Answer 1

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

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

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

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

Answer 2

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 .. . . 

Answer 3

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