=> 3B = 3.( 1/2.5 + 1/5.8 + 1/8.11 + ........... + 1/122.125)
= 3/2.5 + 3/5.8 + 3/ 8.11 + ......+ 3/122.125
Ta có: 3/ 2.5 = 1/2 - 1/5
3/5.8 = 1/5 -1/8
3/ 8.11 = 1/8 -1/11
..........................
3/122 . 125 = 3/122 - 3/125
=> 3B= 1/2 - 15/5 + 1/5 -1/8 +1/8 - 1/11 +........+1/122 - 1/125
= 1/2 - 1/125 = 125/250 - 2/250= 123/250
=> B= 3B : 3 = 123/250 :3 = 123/250 . 1/3 = 41/250
=> 2C = 2.(1/9.11 + 1/11.13 +....+ 1/97 .99)
= 2/9.11 + 2/11 .13 +.....+ 2/ 97.99
Ta có: 2/9.11 = 1/9 - 1/11
2/11.13 = 2/11 -2/ 13
...............................
2/97.99 = 1/97 - 1/99
=> 2B = 1/9 - 1/11 + 1/11 - 1/13 + ....+ 1/97 - 1/99
= 1/9 -1/99 = 11/99 - 1/99 =10/99
=> B= 2B : B = 10/99 :2 =10/99 . 1/2 = 5/99
Vậy B = 5/99
\(B=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{122.125}\)
\(3B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{122.125}\)
Nhận xét:
\(\frac{3}{2.5}=\frac{1}{2}-\frac{1}{5}=\frac{3}{10}\)
\(\frac{3}{5.8}=\frac{1}{5}-\frac{1}{8}=\frac{3}{40}\)
\(\frac{3}{8.11}=\frac{1}{8}-\frac{1}{11}=\frac{3}{88}\)
.............
Từ nhận xét trên ta có:
\(3B=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{122}-\frac{1}{125}\)
\(3B=\frac{1}{2}-\frac{1}{125}=\frac{123}{250}\)
\(B=\frac{123}{250}:3=\frac{41}{250}\)
\(C=\frac{1}{9.11}+\frac{1}{11.13}+...+\frac{1}{97.99}\)
\(2C=\frac{2}{9.11}+\frac{2}{11.13}+...+\frac{2}{97.99}\)
Nhận xét:
\(\frac{2}{9.11}=\frac{1}{9}-\frac{1}{11}=\frac{2}{99}\)
\(\frac{2}{11.13}=\frac{1}{11}-\frac{1}{13}=\frac{1}{143}\)
..................
Từ nhận xét trên ta có:
\(2C=\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{99}\)
\(2C=\frac{1}{9}-\frac{1}{99}=\frac{10}{99}\)
\(C=\frac{10}{99}:2=\frac{5}{99}\)
