x/1x2 +x/2x3 + x/3x4+ .....+x/2017x2018=-1
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(C=1-\frac{1}{2018}\)
\(C=\frac{2017}{2018}\)
\(C=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+.....+\frac{1}{2017x2018}\)
Ta thấy \(\frac{1}{1x2}=\frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{2x3}=\frac{1}{2}-\frac{1}{3}\)
.............................................
\(\frac{1}{2017x2018}=\frac{1}{2017}-\frac{1}{2018}\)
\(\Rightarrow C=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2017}-\frac{1}{2018}\)
\(\Rightarrow C=\frac{1}{1}-\frac{1}{2018}\)
\(\Rightarrow C=\frac{2017}{2018}\)
Chúc bạn học tốt nhớ k mình nhá
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(S=1-\frac{1}{2018}\)
\(S=\frac{2018}{2018}-\frac{1}{2018}\)
\(S=\frac{2017}{2018}\)
\(S=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}.\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{2017}+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}=\frac{2017}{2018}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có công thức tổng quát là:
\(\frac{n.\left(n+1\right).\left(2n+1\right)}{6}\)
Thay vào sẽ là:
\(\frac{2017.2018.\left(2.2017+1\right)}{6}=2737280785\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi B = 1x2 + 2 x 3 + 3 x 4 + ... + 2016 x2017
3B = 3 x ( 1x2 + 2x3 + 3x4 + ... + 2016x2017)
= 1x2x3 + 2x3x3 + 3x4x3 + ... + 2016x2017x3 )
= 1x2x3 + 2x3x( 4-1) + 3x4x( 5 -2 ) + ... + 2016x2017x( 2018 - 2015)
= 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + ... + 2016x2017x2018 - 2015x2016x2017
= 2016 x2017 x2018
B = 672 x2017 x2018
Mà A = \(\frac{672x2017x2018}{2017x2018}\)
= 672
Vậy A = 672
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)
\(=1+\left(1-\frac{1}{2018}\right)\)
\(=1+\left(\frac{2018}{2018}-\frac{1}{2018}\right)\)
\(=1+\left(\frac{2017}{2018}\right)\)
\(=\frac{2018}{2018}+\frac{2017}{2018}=\frac{4035}{2018}\)
\(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}...+\frac{1}{2017\cdot2018}\)
\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}...+\frac{1}{2017}-\frac{1}{2018}\right)\)
\(=1+\left(1-\frac{1}{2018}\right)\)
\(=1+\frac{2017}{2018}\)
\(=1+\frac{2017}{2018}\)
\(=\frac{4035}{2018}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
P=1x2+2x3+3x4+...+2017x2018
3P = 1x2x3 + 2x3x3 + 3x4x3 + ... + 2017x2018x3
3P = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + ... +2017x2018x(2019-2016)
3P = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + ... + 2017x2018x2019 - 2016x2017x2018
3P = 2017x2018x2019
P = 2017x2018x2019 : 3
P = 2739315938
P = 1x2+2x3+3x4+...+2017x2018
3xP = 1x2x3+2x3x3+3x4x3+...+2017x2018x3
3xP = 1x2x3+2x3x(4-1)+3x4x(5-2)+...+2017x2018x(2019-2016)
3xP = 1x2x3+2x3x4-2x3x1+3x4x5-3x4x2+...+2017x2018x2019-2017x2018x2016
3xP = 2017x2018x2019
3xP = 8217947814
P = 8217947814 : 3
P = 2739315938
![](https://rs.olm.vn/images/avt/0.png?1311)
Đề đã đầy đủ chưa bạn? Và bạn đang cần làm gì với biểu thức này?
x= -2018/2017
Bài làm:
Ta có: \(\frac{x}{1.2}+\frac{x}{2.3}+\frac{x}{3.4}+...+\frac{x}{2017.2018}=-1\)
\(\Leftrightarrow x\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\right)=-1\)
\(\Leftrightarrow x\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)=-1\)
\(\Leftrightarrow x\left(1-\frac{1}{2018}\right)=-1\)
\(\Leftrightarrow x.\frac{2017}{2018}=-1\)
\(\Rightarrow x=-\frac{2018}{2017}\)