1)\(S=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{2003.2005.2007}\)

\(\Rightarrow4S=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{2003.2005.2007}\)

\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{2003.2005}-\frac{1}{2005.2007}\)

\(=\frac{1}{3}-\frac{1}{4024035}=\frac{1341345}{4024035}=\frac{1}{3}\)

\(\Rightarrow S=\frac{1}{3}:4\approx0,08\)

2)\(S=\frac{1}{3}:4=\frac{1}{12}\)

\(S=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{2003.2005.2007}\)

\(S=\frac{2}{2}.\frac{1}{1.3.5}+\frac{2}{2}.\frac{1}{3.5.7}+\frac{2}{2}.\frac{1}{5.7.9}+...+\frac{2}{2}.\frac{1}{2003.2005.2007}\)

\(S=\frac{1}{2}.\frac{2}{1.3.5}+\frac{1}{2}.\frac{2}{3.5.7}+\frac{1}{2}.\frac{2}{5.7.9}+...+\frac{1}{2}.\frac{2}{2003.2005.2007}\)

\(S=\frac{1}{2}.\left(\frac{2}{1.3.5}+\frac{2}{3.5.7}+\frac{2}{5.7.9}+...+\frac{2}{2003.2005.2007}\right)\)

\(S=\frac{1}{2}.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{2003.2005}-\frac{1}{2005.2007}\right)\)

\(S=\frac{1}{2}\left(\frac{1}{1.3}-\frac{1}{2005.2007}\right)=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{4024035}\right)=\frac{1}{2}.\frac{1341345}{4024035}=\frac{1}{2}.\frac{1}{3}=\frac{1}{6}\)

Vậy \(S=\frac{1}{6}\)

Đặt A = 1.3.5 + 3.5.7 + 5.7.9 + ...... + 47.49.51

8A = 1.3.5.8 + 3.5.7.8 + ...... + 47.49.51.8

= 1.3.5(7 + 1) + 3.5.7.(9 - 1) + ...... + 47.49.51(53 - 45)

= 1.3.5.7 + 1.3.5 + 3.5.7.9 - 1.3.5.7 + ......... + 47.49.51.53 - 47.47.49.51

= 1.3.5 + 47.49.51.53

=> A = \(\frac{1.3.5+47.49.51.53}{8}=778128\)

628414

chi tiết là ko biết bởi vì e mới học lớp 5

Các bạn giải nhanh lên nhé, mai mình phải nộp cô rồi

1.3.5+3.5.7+5.7.9+...+97.99.101

=(101-2).(101-1).101.(101+1):4

=25497450

1/1.3.5 + 1/3.5.7 + 1/5.7.9 +.....+ 1/99.101.103

= 1/4. [4/1.3.5 + 4/3.5.7 + 4/ 5.7.9 +....+ 4/99.101.103]

=1/4. [1/1.3 - 1/3.5 + 1/3.5 - 1/5.7 +....+ 1/99.101 - 1/101.103]

= 1/4. [1/1.3 - 1/101.103]

=1/4. 10406/31209

= 5230/62418

\(A=\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+....+\frac{1}{99\cdot101\cdot103}\)

\(2A=\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5-7}+....+\frac{1}{99\cdot101}-\frac{1}{101\cdot103}\)

\(2A=\frac{1}{1\cdot3}-\frac{1}{101\cdot103}\)

Tính nốt

Gọi B là biểu thức đã cho.
Dễ dàng chứng minh:

...
Ta có:
B