K
Khách

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

27 tháng 2 2017

Chào bạn !Mình kết bạn nha!

27 tháng 2 2017
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                               
\(B=\frac{3+1}{3}+\frac{8+1}{8}+...+\frac{2014.2016+1}{2014.2016}+\frac{2015.2017+1}{2015.2017}\)         
\(=\frac{2^2}{3}+\frac{3^2}{8}+....+\frac{2015^2}{2014.2016}+\frac{2016^2}{2015.2017}\)         
=\(\frac{2.2}{3}+\frac{3.3}{2.4}+...+\frac{2015.2015}{2014.2016}+\frac{2016.2016}{2015.2017}\)         
=\(\frac{\left(2.3....2015.2016\right)+\left(2.3.....2015.2016\right)}{\left(1.2.3.....2014.2015\right)+\left(3.4....2016.2017\right)}\)         
=\(2016+\frac{2}{2017}\)         
          
          
          
          
          
          
          
          
          
          
          
                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
 
12 tháng 5 2017

Lê Châu bị Điên à

12 tháng 5 2017

lê châu bị khùng à

6 tháng 4 2022

\(C=\left(1+\frac{1}{1.3}\right)\)\(.\left(1+\frac{1}{2.4}\right)\)\(.\left(1+\frac{1}{3.5}\right)\)\(.\left(1+\frac{1}{2014.2016}\right)\)

   \(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{2015^2}{2014.2016}\)

   \(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{2015.2015}{2014.2016}\)

   \(=\frac{\left(2.3.4...2015\right).\left(2.3.4...2015\right)}{\left(1.2.3...2014\right).\left(3.4.5...2016\right)}\)

   \(=\frac{2015.2}{2016}\)

    \(=...\)(tự tinhs)

12 tháng 3 2016

giúp mới mình sẽ tích

12 tháng 3 2016

bn viết rõ đề đi

10 tháng 1 2016

5435

tích mình nha

10 tháng 1 2016

luu y : dau /la phan cach giua mau so va tu so

16 tháng 4 2016

ket qua la 5435 ban nha

19 tháng 3 2017

Ta có công thức :

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

Áp dụng vào bài toán ta được :

\(C=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}..........\frac{2015^2}{2014.2016}\)

\(=\frac{\left(2.3.4....2015\right)\left(2.3.4....2015\right)}{\left(1.2.3...2014\right)\left(3.4.5.....2016\right)}\)

\(=\frac{2015.2}{2016}=\frac{2015}{1008}\)

19 tháng 3 2017

=1(1/1*3*(1/2*4)*...*(1+1/2014*2016)

=1/2(2+2/1*3)+(2+2/2*4)*...(2+2/2014*2016)

=1/2(2+1/1-1/3)...(2+1/2014-1/2016)

=1/2*(1/1-1/2016)

=3023/4032