K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

8 tháng 2 2019

\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}.\frac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}\)

                                      \(=\frac{1}{2}\left[\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}\right]\)

                                      \(=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\)

10 tháng 10 2015

 Xét vế phải: \(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)

= \(\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}\)

= \(\frac{n+2-n}{n\left(n+1\right)\left(n+2\right)}\)

= \(\frac{2}{n\left(n+1\right)\left(n+2\right)}\)   

= VT

=> Đpcm
 

10 tháng 10 2015

quy đồng là ra ngay đó mà

27 tháng 8 2016

\(\frac{1}{2}\left(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)=\frac{1}{2}\left(\frac{\left(n+1\right)\left(n+1\right)}{n\left(n+1\right)\left(n+1\right)\left(n+2\right)}-\frac{n\left(n+1\right)}{n\left(n+1\right)\left(n+1\right)\left(n+2\right)}\right)\)

\(\frac{1}{2}\left(\frac{1}{n\left(n+2\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)=\frac{\left(n+1\right)\left(n+2\right)}{n\left(n+2\right)\left(n+1\right)\left(n+2\right)}-\frac{\left(n+1\right)\left(n+2\right)}{n\left(n+2\right)\left(n+1\right)\left(n+2\right)}\)

\(=\frac{1}{n\left(n+1\right)\left(n+2\right)}\)

 

27 tháng 8 2016

\(\frac{1}{2}\left(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)=\frac{1}{2}\left(\frac{\left(n+1\right)\left(n+1\right)}{n\left(n+1\right)\left(n+1\right)\left(n+2\right)}-\frac{n\left(n+1\right)}{n\left(n+1\right)\left(n+1\right)\left(n+2\right)}\right)\)

\(=\frac{1}{2}\left(\frac{1}{n\left(n+2\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)=\frac{\left(n+1\right)\left(n+2\right)}{n\left(n+2\right)\left(n+1\right)\left(n+2\right)}-\frac{\left(n+1\right)\left(n+2\right)}{n\left(n+2\right)\left(n+1\right)\left(n+2\right)}\)

\(=\frac{1}{n\left(n+1\right)\left(n+2\right)}\)

CM : \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\)

Có : \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}.\frac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}\)

\(\frac{1}{n\left(n+1\right)\left(n+2\right)}\)\(=\frac{1}{2}\left[\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}\right]\)

\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\) đpcm

15 tháng 6 2019

Cảm ơn bạn

2 tháng 4 2018

Ta có : \(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}=\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}=\frac{2}{n\left(n+1\right)\left(n+2\right)}\)

Vì VT=VP nên ta có đpcm

2 tháng 4 2018

\(\text{Ta có:}\)

\(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}=\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n+1}+\frac{1}{n+2}=\frac{2\left(n+1\right)}{n\left(n+2\right)}-\frac{2}{n+1}\left(1\right)\)

\(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{2}{n\left(n+1\right)\left(n+2\right)}+\frac{2}{n+1}-\frac{2}{n+1}=\frac{2n\left(n+2\right)+2}{n\left(n+1\right)\left(n+2\right)}-\frac{2}{n+1}=\frac{2\left(n+1\right)^2}{n\left(n+1\right)\left(n+2\right)}-\frac{2}{n+1}=\frac{2\left(n+1\right)}{n\left(n+2\right)}-\frac{2}{n+1}\left(2\right)\)

\(\text{Từ (1) và (2) ta có: ĐPCM}\)

21 tháng 1 2017

Ta có \(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)

\(=\frac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}=\frac{2}{n\left(n+1\right)\left(n+2\right)}\) (đpcm)

Áp dụng công thức trên ta có

A\(=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\cdot\cdot\cdot\cdot\cdot\cdot\cdot+\frac{1}{2015\cdot2016\cdot2017}\)

\(\Leftrightarrow2A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{2015\cdot2016\cdot2017}\)

\(\Leftrightarrow2A=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{2}{3\cdot4}+....+\frac{1}{2015\cdot2016}-\frac{1}{2016\cdot2017}\)

\(\Leftrightarrow2A=\frac{1}{1\cdot2}-\frac{1}{2016\cdot2017}\)

\(\Rightarrow A=\left(\frac{1}{1\cdot2}-\frac{1}{2016\cdot2017}\right)\div2\approx0.25\)

Vậy A\(\approx0.25\)

24 tháng 7 2019

\(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)

\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{3n+2}\right)\)

\(=\frac{1}{3}.\frac{3n}{2.\left(3n+2\right)}\)

\(=\frac{n}{2\left(3n+2\right)}\)