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

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

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

Ta có đpcm. 

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

\(\text{Ta có:}\)

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

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

\(\text{Từ (1) và (2) ta có: ĐPCM}\)

Với \(k\in N;k>0\) Ta có :

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

Áp dụng ta có :

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+.....+\frac{1}{\left(n-1\right)n\left(n+1\right)}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{n\left(n+1\right)}\right)=\frac{1}{2}.\frac{n\left(n+1\right)-2}{2n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{4n\left(n+1\right)}\)(đpcm)

Ta có : 

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\left(n-1\right)n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{4n\left(n+1\right)}\)

\(\Leftrightarrow\)\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{\left(n-1\right)n\left(n+1\right)}=\frac{2\left(n-1\right)\left(n+2\right)}{4n\left(n+1\right)}\)

\(\Leftrightarrow\)\(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}=\frac{n\left(n-1\right)+2\left(n-1\right)}{2n\left(n+1\right)}\)

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

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

\(\Leftrightarrow\)\(\frac{n^2+n-2}{2n^2+2n}=\frac{n^2+n-2}{2n^2+2n}\) với \(n\ge2\)

Vậy ...

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

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

a/  Xét mẫu số VP_  n và n+1 là 2 số liên tiếp 

\(\Rightarrow\left(n,n+1\right)\)bằng 1

Thay vào đề bài     \(\frac{1}{n}-\frac{1}{n+1}\)bằng   \(\frac{n+1}{n.\left(n+1\right)}-\frac{n}{n.\left(n+1\right)}\)bằng \(\frac{1}{n\cdot\left(n+1\right)}\)

\(\Rightarrowđpcm\)

P/s _laptop ko gõ đc dấu

 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
 

quy đồng là ra ngay đó mà

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\(\approx0.25\)