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9 tháng 5 2016

http://olm.vn/hoi-dap/question/161872.html

9 tháng 5 2016

B=1/1.3+1/3.5+...+1/2015.2017

B= 1/2 . 2. ( 1/1.3+1/3.5 + .... + 1/2015 .2017)

B = 1/2 . ( 2/1.3 + 2/3.5 + ......+ 2/2015.2017)

B = 1/2. ( 1/1+ -1/3 + 1/3 + -1/5 + 1/5 +....+ 1/2015 + -1/2017)

B= 1/2 . ( 1/1 + -1 / 2017) = 1/2 . 2016 / 2017 = 2016 / 4034

Vậy B = 2016 / 4034 nha bn.pham hong thai

9 tháng 8 2015

Dễ thôi:

Khoảng cách là 2

\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(\frac{1}{2}.\left(1-\frac{1}{2017}\right)=\frac{1}{2}.\frac{2016}{2017}=\frac{1008}{2017}\)

20 tháng 6 2020

cảm ơn bạn đã giúp mình!!

21 tháng 5 2018

Ta có : \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2015.2017}\)

           \(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2015.2017}\right)\)

           \(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)

            \(=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)

             \(=\frac{1}{2}.\frac{2016}{2017}=\frac{1008}{2017}\)

21 tháng 5 2018

\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2015.2017}\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)

\(=\frac{1}{2}.\frac{2016}{2017}\)

\(=\frac{1008}{2017}\)

21 tháng 8 2016

Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\), ta có:

\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)

\(=\frac{1}{2}.\frac{2016}{2017}=\frac{1008}{2017}\)

21 tháng 8 2016

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2015}-\frac{1}{2017}+\frac{1}{2017}\)

\(=1-\frac{1}{2017}\)

\(=\frac{2016}{2017}\)

mk đầu tiên đấy

27 tháng 2 2017

Chào bạn !Mình kết bạn nha!

27 tháng 2 2017
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                               
\(B=\frac{3+1}{3}+\frac{8+1}{8}+...+\frac{2014.2016+1}{2014.2016}+\frac{2015.2017+1}{2015.2017}\)         
\(=\frac{2^2}{3}+\frac{3^2}{8}+....+\frac{2015^2}{2014.2016}+\frac{2016^2}{2015.2017}\)         
=\(\frac{2.2}{3}+\frac{3.3}{2.4}+...+\frac{2015.2015}{2014.2016}+\frac{2016.2016}{2015.2017}\)         
=\(\frac{\left(2.3....2015.2016\right)+\left(2.3.....2015.2016\right)}{\left(1.2.3.....2014.2015\right)+\left(3.4....2016.2017\right)}\)         
=\(2016+\frac{2}{2017}\)         
          
          
          
          
          
          
          
          
          
          
          
                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
                                                                                                                                                                                                                                      
 
9 tháng 3 2021

\(A=\dfrac{1}{2}\left(2.\dfrac{2}{3}\right)\left(\dfrac{3}{2}.\dfrac{3}{4}\right)\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)

\(=\dfrac{2016}{2017}\)

7 tháng 2 2020

Lời giải:

Xét tổng quát:

1+1k(k+2)=k(k+2)+1k(k+2)=(k+1)2k(k+2)1+1k(k+2)=k(k+2)+1k(k+2)=(k+1)2k(k+2)

Thay k=1,2,....,2015k=1,2,....,2015 ta có:

1+11.3=221.31+11.3=221.3

1+12.4=322.41+12.4=322.4

1+13.5=423.51+13.5=423.5

1+14.6=524.61+14.6=524.6

.............

1+12015.2017=201622015.20171+12015.2017=201622015.2017

Nhân theo vế:

⇒A=12(1+11.3)(1+12.4)(1+13.5)....(1+12015.2017)⇒A=12(1+11.3)(1+12.4)(1+13.5)....(1+12015.2017)

=12.221.3.322.4.423.5.524.6....201622015.2017=12.221.3.322.4.423.5.524.6....201622015.2017

=(1.2.3...2016)2(1.2.3...2015)(2.3.4...2017)=(1.2.3...2016)(2.3....2016)(1.2.3...2015)(2.3.4...2017)=2016.12017=20162017

7 tháng 2 2020

đây là tính nhanh đấy