\(B=\frac{\frac{2016}{1000}+\frac{2016}{999}+...+\frac{2016}{501}}{\frac{-1}{1.2}+\frac{-1}{3.4}+...+\frac{-1}{999.1000}}=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{999.1000}\right)}\)

\(=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}\right)}\)

\(=\frac{2016\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{999}+\frac{1}{1000}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\right)\right]}\)

\(=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{999}+\frac{1}{1000}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{500}\right)\right]}\)

\(=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+....+\frac{1}{999}+\frac{1}{1000}\right)}=\frac{2016}{-1}=-2016\)

Vậy B = - 2016

Bạn Xyz cho mik hỏi ở phần mẫu số tại sao lại có -2*(1/2+1/4+...+1/1000) vậy? Nó ở đâu ra thế?

\(B=\frac{\frac{2016}{1000}+\frac{2016}{999}+\frac{2016}{998}+...+\frac{2016}{501}}{-\frac{1}{1\cdot2}-\frac{1}{3\cdot4}-\frac{1}{5\cdot6}-...-\frac{1}{999\cdot1000}}\)

\(B=\frac{2016\left(\frac{1}{1000}+\frac{1}{999}+\frac{1}{998}+...+\frac{1}{501}\right)}{-\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{999\cdot1000}\right)}\)

\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{999}-\frac{1}{1000}\right)}\)

\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left[\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\right)\right]}\)

\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1000}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\right)\right]}\)

\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1000}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{500}\right)}\)

\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}\)

\(B=\frac{2016}{-1}=-2016\)

cảm ơn bạn Phương Uyên

B = 1/1.2 + 1/3.4 +..+1/2015.2016
B = 1-1/2 + 1/3 - 1/4 +...+ 1/2015 - 1/2016
B = 1+ 1/2 + 1/3 +..+1/2015 + 1/2016 - 2( 1/2 + 1/4 + ..1/2016)
B = 1/1009 + 1/1010 +.. + 1/2016 ( dpcm)

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2015.2016}\)

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2015}-\frac{1}{2016}\)

\(A=\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{1008}\right)\)

\(A=\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2015}+\frac{1}{2016}\)

\(\Rightarrow B-A=\left(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)-\left(\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2016}\right)\)

\(\Rightarrow B-A=\frac{1}{1008}\)