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Xét số hạng một cách tổng quát:
1+1/[k.(k+2)]=[k.(k+2)+1]/[k.(k+2)]=(k^2+2k+1)/[k.(k+2)]=(k+1)^2/[k.(k+2)]
Cho k đi từ 1 đến 2018 ta sẽ có:
*1+1/1.3=2^2/1.3
*1+1/2.4=3^2/2.4
*1+1/3.5=4^2/3.5
..................
*1+1/2016.2018=2017^2/2016.2018
*1+1/2017.2019=2018^2/2017.2019
*1+1/2018.2020=2019^2/2018.2020
Ta thay vào B = ( 1 + 1/1.3 ) . ( 1 + 1/2.4 ) + ( 1 + 1/3.5 ) + .....+ ( 1 + 1/2018.2020 )
=[2^2.3^2...2019^2]/[1.2.3^2.4^2.5^2.6^2...2018^2.2019.2020]
=[2^2.2019^2]/(2.2019.2020]
=2.2019/2020
=4038/2020
B= (1*3+1/1*3)*(2*4+1/2*4)*....*(2018*2020+1/2018*2020)
B=(4/1*3)*(9/2*4)*...*(4076361/2018*2020)
B=(2*2/1*3)*(3*3/2*4)*...*(2019*2019/2018*2020)
B=(2*3*...*2019)*(2*3*...*2019)/(1*2*...*2018)*(3*4*...*2020)
B=2019/2020
nhớ cho mình 1 k và kết bạn nhé
\(B=2016.\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2014.2016}\right)\)
= \(2016.\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}....\frac{2015^2}{2014.2016}\)
= \(2016.\frac{2.3.4....2015}{1.2.3.4.5...2014.2015.2016}.\frac{2.3.4....2015}{3.4.5...2014}\)
= \(2016.\frac{1}{2016}.2.2015=2.2015=4030\)
= 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
\(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)
\(\Rightarrow A=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{99.101+1}{99.101}\)
\(\Rightarrow A=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.....\frac{10000}{99.101}\)
\(\Rightarrow A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{100^2}{99.101}\)
\(\Rightarrow A=\frac{\left(2.3.4.....100\right)\left(2.3.4.....100\right)}{\left(1.2.3.....99\right)\left(3.4.5.....101\right)}\)
\(\Rightarrow A=\frac{100.2}{101}=\frac{200}{101}\)
\(=\frac{4}{3}.\frac{9}{8}...\frac{4060225}{4060224}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}...\frac{2015.2015}{2014.2016}\)
\(=\frac{2.2.3.3...2015.2015}{1.3.2.4...2014.2016}\)
\(=\frac{2.3...2015}{1.2...2014}.\frac{2.3...2015}{3.4...2016}\)
\(=2015.\frac{2}{2016}\)
\(=2015.\frac{1}{1008}\)
\(=\frac{2015}{1008}\)
Ta có
=\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right)....\left(1+\frac{1}{8.10}\right)\)
=\(\frac{4}{3}.\frac{9}{8}....\frac{81}{80}\)
=\(\frac{2.2}{1.3}.\frac{3.3}{2.4}....\frac{9.9}{8.10}\)
=\(\frac{2.3....9}{1.2....8}.\frac{2.3....9}{3.4....10}\)
=\(9.\frac{2}{10}\)
=\(\frac{9}{5}\)
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.
\(B=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right).....\left(1+\frac{1}{2016.2018}\right)\)
\(=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{2016.2018+1}{2016.2018}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{2017^2}{2016.2018}\)
\(=\frac{\left(2.3.4.....2017\right)\left(2.3.4.....2017\right)}{\left(1.2.3.....2016\right)\left(3.4.5.....2018\right)}\)
\(\Rightarrow B=\frac{2017.2}{2018.1}=\frac{4034}{2018}=\frac{2017}{1009}\)