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14 tháng 2 2018

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)

14 tháng 2 2018

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 ...

24 tháng 12 2016

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

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

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

\(2A=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\Rightarrow A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)

mình áp dụng công thức tổng quát:\(\frac{a}{n\left(n+1\right)\left(n+2\right)...\left(n+a\right)}=\frac{1}{n\left(n+1\right)\left(n+a-1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)...\left(n+a\right)}\)

hihi

24 tháng 12 2016

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

<=>\(2A=2\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\right)\)

<=>\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)

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

<=>\(2A=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}=\frac{\left(n+1\right)\left(n+2\right)-2}{2\left(n+1\right)\left(n+2\right)}=\frac{n^2+3n}{2\left(n+1\right)\left(n+2\right)}=\frac{n\left(n+3\right)}{2\left(n+1\right)\left(n+2\right)}\)

<=>\(A=\frac{n\left(n+3\right)}{2\left(n+1\right)\left(n+2\right)}.\frac{1}{2}=\frac{n\left(n+3\right)}{4\left(n+1\right)\left(n+2\right)}\)

24 tháng 12 2016

tổng quát:  1/n(n+1)(n+2)=1/2[1/n(n+1) - 1/(n+1)(n+2)]

18 tháng 1 2018

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

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

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

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

P/S:  tham khảo nhé

đến đây bn làm tiếp nha

26 tháng 2 2017

\(\frac{150}{5.8}+\frac{150}{8.11}+\frac{150}{11.14}+.....+\frac{150}{47.50}\)

\(=50.\left(\frac{3}{5.8}+\frac{5}{8.11}+.....+\frac{3}{47.50}\right)\)

\(=50.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{47}-\frac{1}{50}\right)\)

\(=50.\left(\frac{1}{5}-\frac{1}{50}\right)\)

\(=50.\frac{9}{50}=9\)

1 tháng 12 2016

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)

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

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

\(=\frac{1}{4}-\frac{1}{2\left(n+1\right)\left(n+2\right)}\) \(< \frac{1}{4}\)

7 tháng 5 2017

bài này tớ chịu!

2 tháng 11 2019

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

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\)

\(A=1-\frac{1}{n+1}\)

2 tháng 11 2019

a) Ta có: \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\)

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

           \(A=1-\frac{1}{n+1}\)

           \(A=\frac{n+1}{n+1}-\frac{1}{n+1}\)

           \(A=\frac{n}{n+1}\)

Học tốt nha^^

13 tháng 6 2016

Tất cả các đẳng thức trên đều được chứng minh theo phương pháp quy nạp

Đặt n = k thì có đẳng thức

Chứng minh rằng n = k+1 cũng đúng ( vế trái (k+1) = vế phải (k+1) )

13 tháng 6 2016

thi giai ra luon dj

15 tháng 7 2018

Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)

   \(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)

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

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

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

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

\(\Rightarrow A=\frac{\left(n+1\right)\left(n+2\right)-2}{4\left(n+1\right)\left(n+2\right)}\)

TK nha!!