Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=124\left(\frac{1}{1.1985}+\frac{1}{2.1986}+\frac{1}{3.1987}+...+\frac{1}{16.2000}\right)\)
\(=\frac{124}{1984}.\left(1-\frac{1}{1985}+\frac{1}{2}-\frac{1}{1986}+...+\frac{1}{16}-\frac{1}{2000}\right)\)
\(=\frac{1}{16}\left[\left(1+\frac{1}{2}+...+\frac{1}{16}\right)-\left(\frac{1}{1985}+\frac{1}{1986}+...+\frac{1}{2000}\right)\right]\)
Và \(B=\frac{1}{1.17}+\frac{1}{2.18}+...+\frac{1}{1984.2000}\)
\(=\frac{1}{16}\left[\left(1-\frac{1}{17}+\frac{1}{2}-\frac{1}{18}+...+\frac{1}{1984}-\frac{1}{2000}\right)\right]\)
\(=\frac{1}{16}\left[\left(1+\frac{1}{2}+...+\frac{1}{1984}\right)-\left(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{2000}\right)\right]\)
= \(\frac{1}{16}\) . \(\left[\left(1+...+\frac{1}{16}\right)+\left(\frac{1}{17}+...+\frac{1}{1984}-\frac{1}{17}-...-\frac{1}{1984}\right)-\left(\frac{1}{1985}+...+\frac{1}{2000}\right)\right]\)
= \(=\frac{1}{16}\left[\left(1+\frac{1}{2}+...+\frac{1}{16}\right)-\left(\frac{1}{1985}+\frac{1}{1986}+...+\frac{1}{2000}\right)\right]\)
Vậy A = B
Ok
A=124.(1/1985+1/2.1986+...+1/16.2000)
A=124/1984.(1/1-1/1985)+(1/2-1/1986)+...+(1/16-1/2000)
A=1/16.(1+1/2+1/3+...+1/16)-(1/1985+1/1986+...+1/2000) (1)
B=1/1.17+1/2.18+1/3.19+...+1/1984.2000
B=1/16.(1-1/7+1/2-1/18+1/3-1/19+...+1/1984-1/2000)
B=1/16.(1-1/2-1/3+...+1/1984)-(1/17+1/18+1/19+...+1/2000)
B=1/16.(1/1+1/2+1/3+...+1/16)+(1/17+1/18+...+1/1984)-(1/17+1/18++...+1/1984)-(1/1985+1/1986+...+1/2000)
B=1/16.(1+1/2+...+1/8)-(1/1985+...+1/2000) (2)
Từ (1) và (2)
452