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14 tháng 3 2017

kq :2015/6054

7 tháng 3 2018

\(S=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2016.2018}\right)\)

\(\Rightarrow S=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{2016.2018+1}{2016.2018}\)

\(\Rightarrow S=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{2017^2}{2016.2018}\)

\(\Rightarrow S=\frac{\left(2.3.4.....2017\right)\left(2.3.4.....2017\right)}{\left(1.2.3.....2016\right)\left(3.4.5.....2018\right)}\)

\(\Rightarrow S=\frac{2017.2}{1.2018}=\frac{4034}{2018}=\frac{2017}{1009}\)

Tự chứng minh: 1 + 1/n(n+2)=(n+1)2/n(n+2)

Áp dụng đẳng thức trên, ta có: 

1 + 1/1.3= 22/1.3

1 + 1/2.4= 32/2.4

...

1 + 1/2016.2018=20172/2016.2018

Đến đó tự làm nha bạn, máy mình không bấm được phân số, thông cảm.

27 tháng 3 2019

\(B=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right).....\left(1+\frac{1}{2016.2018}\right)\)

\(=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{2016.2018+1}{2016.2018}\)

\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{2017^2}{2016.2018}\)

\(=\frac{\left(2.3.4.....2017\right)\left(2.3.4.....2017\right)}{\left(1.2.3.....2016\right)\left(3.4.5.....2018\right)}\)

\(\Rightarrow B=\frac{2017.2}{2018.1}=\frac{4034}{2018}=\frac{2017}{1009}\)

hơi khó đó tick mình nha Hoàng Thu Hà

4 tháng 11 2016

\(\frac{1}{2}.\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2015.2017}\right)\)

\(=\frac{1}{2}.\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}...\frac{2015.2017+1}{2015.2017}\)

\(=\frac{1}{2}.\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{2016.2016}{2015.2017}\)

\(=\frac{1}{2}.\frac{2.3.4...2016}{1.2.3...2015}.\frac{2.3.4...2016}{3.4.5...2017}\)

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

28 tháng 12 2016

tuyệt

6 tháng 7 2017

= 4/1.3 x 9/2.4 x 16/3.5 x...x 10000/99.101

= 2.2/1.3 x 3.3/2.4 x 4.4/3.5 x..x 100.100/99.101

= (2.3.4. ... 100/1.2.3. .... 99) x (2.3.4. ... .100/3.4.5. ... .101)

= 100.2/101

=200/101

7 tháng 3 2018

\(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)

\(\Rightarrow A=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{99.101+1}{99.101}\)

\(\Rightarrow A=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.....\frac{10000}{99.101}\)

\(\Rightarrow A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{100^2}{99.101}\)

\(\Rightarrow A=\frac{\left(2.3.4.....100\right)\left(2.3.4.....100\right)}{\left(1.2.3.....99\right)\left(3.4.5.....101\right)}\)

\(\Rightarrow A=\frac{100.2}{101}=\frac{200}{101}\)

4 tháng 11 2016

\(\frac{1}{2}.\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2015.2017}\right)\)

\(=\frac{1}{2}.\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}...\frac{2015.2017+1}{2015.2017}\)

\(=\frac{1}{2}.\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{2016.2016}{2015.2017}\)

\(=\frac{1}{2}.\frac{2.3.4...2016}{1.2.3...2015}.\frac{2.3.4...2016}{3.4.5...2017}\)

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