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

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

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

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

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

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?

= (1/1 -1/2) + (1/2-1/3) + (1/3x1/4)+...+(1/999- 1/1000)

= 1/1- 1/1000

= ...[bn tự tính nhé]

k mk nha, nếu đúng

\(\text{Đề bạn bị sai thì phải ????? Đề đúng phải là }:\)

      \(\frac{1}{1\text{ x }2}+\frac{1}{2\text{ x }3}+\frac{1}{3\text{ x }4}+...+\frac{1}{999\text{ x }1000}\)

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

\(=1-\frac{1}{1000}\)

\(=\frac{999}{1000}\)

1/1x2+1/2x3+1/3x4+...

= 1/1-1/2+1/2-1/3+1/3-1/4+...

= 1-1/4

=3/4

K nhé

bạn trả lời đúng rồi nhưng giúp mình phần sau nữa nhé !

ARIGATO

1/1x2+1/2x3+...+1/999x1000+1

= 1+ (1/1-1/2+1/2-1/3+...+1/999-1/1000)

= 1 + ( 1-1/1000)

= 1 + 999/1000

= 1999/1000

K mk nhé bn

=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+........+\frac{1}{999}-\frac{1}{1000}+1\)

=\(\frac{1}{1}-\frac{1}{1000}+1\)

=\(\frac{1999}{1000}\)

     \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{999\cdot1000}+1\)

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

= \(1-\frac{1}{1000}+1\)

= \(\frac{999}{1000}+1\)

=\(\frac{1999}{1000}\)

                                                     \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{999\cdot1000}+1\)

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

                                                   = \(\frac{1}{1}-\frac{1}{1000}+1\)

                                                   = \(\frac{999}{1000}+1\)

                                                   = \(\frac{999}{1000}+\frac{1000}{1000}\) 

                                                   = \(\frac{1999}{1000}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}+1\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}+1\)

\(=1-\frac{1}{1000}+1=\frac{1999}{1000}\)

1/2*2+1/2*3+...+1/999*1000+1

=1-1/2+1/2-1/3+1/3-1/4+...+1/999-1/1000+1

=1+1-1/1000

=1999/1000

1/1 - 1/1000 + 1 = 1999/1000

đúng 100% đó bạn , nha

\(\frac{1}{1x2}+\frac{1}{2x3}+.................+\frac{1}{999.1000}+1\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.............+\frac{1}{999}-\frac{1}{1000}+1\)

\(=1-\frac{1}{1000}+1\)

\(=\frac{999}{1000}+1\)

\(=\frac{1999}{1000}\)