1/1.2 + 1/2.3 + 1/3.4+.....+1/1999.2000 = ?
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=2666666000
Có công thức như sau
1x2+2x3+3x4+...+nx(n+1)=nx(n+1)x(n+2):3
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2000}}{\frac{1999}{1}+\frac{1998}{2}+....+\frac{1}{1999}}\)
Xét mẫu số:
\(\frac{1999}{1}+\frac{1998}{2}+\frac{1997}{3}+\frac{1996}{4}+....+\frac{1}{1999}\)
=\(\left(\frac{1998}{2}+1\right)+\left(\frac{1997}{3}+1\right)+\left(\frac{1996}{4}+1\right)+....+\left(\frac{1}{1999}+1\right)+1\)
=\(\frac{2000}{2}+\frac{2000}{3}+\frac{2000}{4}+....+\frac{2000}{1999}+\frac{2000}{2000}\)
= 2000\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{1999}+\frac{1}{2000}\right)\)
=> A = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2000}}{2000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2000}\right)}\)
=> A = \(\frac{1}{2000}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có Đặt B = \(\frac{1999}{1}+\frac{1998}{2}+...+\frac{1}{1999}\)(1999 số hạng)
\(=\left(1+1+1+...+1\right)+\frac{1998}{2}+\frac{1997}{3}+...+\frac{1}{1999}\)(1999 số hạng 1)
\(=1+\left(\frac{1998}{2}+1\right)+\left(\frac{1997}{3}+1\right)+...+\left(\frac{1}{1999}+1\right)\)(1998 cặp số)
= \(\frac{2000}{2}+\frac{2000}{3}+...+\frac{2000}{1999}+\frac{2000}{2000}\)
= \(2000\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1999}+\frac{1}{2000}\right)\)
Khi đó \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{\frac{1999}{1}+\frac{1998}{2}+...+\frac{1}{1999}}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{2000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}\right)}=\frac{1}{2000}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1, S1 = (-2) + (-2) +..+ (-2).
Có SS (-2) là :
(1997 - 1) : 4 +1 = 500 (số ).
Tổng số (-2) là: 500 x (-2) = (-1000)
Bạn chờ mình làm tiếp nha
Các bạn ơi làm giúp mình vs ạ,mình đang cần gấp lắm rồi!!!!HELP MEEEEEEEEEEEEEE
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{1999}\right)\left(1-\frac{1}{2000}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{1998}{1999}.\frac{1999}{2000}=\frac{1.2.3...1998.1999}{2.3.4...1999.2000}=\frac{1}{2000}\)
\(\left(1-\frac{1}{2}\right).\left(1.\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{1999}\right).\left(1-\frac{1}{2000}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{1998}{1999}.\frac{1999}{2000}\)
\(=1.\frac{1}{2000}\)
\(=\frac{1}{2000}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{1999}\right).\left(1-\frac{1}{2000}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{1998}{1999}.\frac{1999}{2000}\)(Rút gọn trên tử với dưới mẫu nhé)
\(=\frac{1}{2000}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{A}{B}=\frac{\frac{2000}{1}+\frac{1999}{2}+\frac{1998}{3}+...+\frac{1}{2000}+2000}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)
\(\Leftrightarrow\frac{A}{B}=\frac{\left(\frac{2000}{1}+1\right)+\left(\frac{1999}{2}+1\right)+\left(\frac{1998}{3}+1\right)+...+\left(\frac{1}{2000}+1\right)+2000+1}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)
\(\Leftrightarrow\frac{A}{B}=\frac{\frac{2001}{1}+\frac{2001}{2}+\frac{2001}{3}+...+\frac{2001}{2000}+2001}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)
\(\Leftrightarrow\frac{A}{B}=\frac{2001\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}\right)}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)
\(\Leftrightarrow\frac{A}{B}=2001\)
bn cộng trên tử rồi thì phải trừ đi chứ ko phân số sẽ thay đổi
1/1.2 + 1/2.3 + 1/3.4 + ... + 1/1999.2000
= 1 -1/2+1/2-1/3+1/3-1/4+....+1/1999-1/2000
= 1- 1/2000
= 1999/2000