Đáp án của tớ là:

\(\frac{1}{1002}+\frac{1}{1003}+...+\frac{1}{2003}=\)\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2003}\right)-\)\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1001}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2003}\right)-\)\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2002}\right)-\)\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2002}\right)=\)\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2003}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-...-\frac{1}{2002}\)\(-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-...-\frac{1}{2002}\)

Vậy:\(1+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2003}=\frac{1}{1002}+\frac{1}{1003}+...+\frac{1}{2003}\)

xin chòa hôm nay mình sẽ giúp bạn lam bài toán này 

ta có

1/1002+1/1003+....+1/2003=(1+1/2+1/3+.....+1/2003)-(1+1/2+1/3+....+1/1001)

1/1002+1/1003+....+1/2003=(1+1/2+1/3+.....+1/2003)-(1/2+1/4+1/6+....+1/2002)-(1/2+1/4+1/6+......+1/2002)

1/1002+1/1003+.....+1/2003=1+1/2+1/3+....+1/2003-1/2+1/4+1/6+....+1/2002-1/2-1/4-1/6-....-1/2002

Vậy1/1002+1/1002+.....+1/2003=1-1/2+1/3-1/4+....-2/2002-1/2003

Sửa: Vậy: \(1-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{2003}=\frac{1}{1002}+\frac{1}{1003}+...+\frac{1}{2003}\)

  ta chuyển đề bài vế trái thành:

  (1+1/2+1/3+1/4+...+1/2001+1/2002) - 2(1/2+1/4+1/6+...+1/2002)

=(1+1/2+1/3+....+1/2002) - (1+1/2+1/3+1/4+...+1/1001)

=1/1002+1/1003+...+1/2002

=> điều phải chứng minh

ta chuyển đề bài vế trái thành:

(1+1/2+1/3+1/4+...+1/2001+1/2002) - 2(1/2+1/4+1/6+...+1/2002)

=(1+1/2+1/3+....+1/2002) - (1+1/2+1/3+1/4+...+1/1001)

=1/1002+1/1003+...+1/2002

=> điều phải chứng minh

vì \(\frac{1}{2^2}< \frac{1}{1\cdot2}\)(do 2²  > 1.2)

            \(\frac{1}{3^2}< \frac{1}{2.3}\)(do 3²>2.3)

             \(\frac{1}{4^2}< \frac{1}{3.4}\)(do 4² >3.4)

          ...

           \(\frac{1}{2002^2}< \frac{1}{2001.2002}\)(do 2002² > 2001.2002)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2002^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2001.2002}\)(2)

Ta có : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2001.2002}\)

   \(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2001}-\frac{1}{2002}\)

   \(=\frac{1}{1}-\frac{1}{2002}\) 

    \(=\frac{2002}{2002}-\frac{1}{2002}\)

     \(=\frac{2001}{2002}< 1\)(2)

Từ (1) và (2) suy ra: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2002^2}< 1\)

Bài toán được chứng minh

Lời giải:

Đề sai, đoạn cuối phải là $2001+(-2002)+2003$

$1+(-2)+3+(-4)+....+2001+(-2002)+2003$

$=[1+(-2)]+[3+(-4)]+...+[2001+(-2002)]+2003$

$=\underbrace{(-1)+(-1)+(-1)+...+(-1)}_{1001}+2003$

$=(-1).1001+2003=-1001+2003=1002$ 

Đáp án D.

S=\(\left(1+\frac{1}{2}+......+\frac{1}{2002}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+..........+\frac{1}{2002}\right)\)

=\(\left(1+\frac{1}{2}+.........+\frac{1}{2002}\right)-\left(1+\frac{1}{2}+.........+\frac{1}{1001}\right)\)

=\(\frac{1}{1002}+\frac{1}{1003}+...........+\frac{1}{2002}=P\)

\(\Rightarrow S-P=0\)

làm ơn tách ra giùm mk

ta thấy : \(\dfrac{-1003}{-2002}\) = \(\dfrac{1003}{2002}\)

\(\dfrac{1004}{-2003}\) = \(\dfrac{-1004}{2003}\)

Sắp xếp : \(\dfrac{1004}{-2003}\) <\(\dfrac{-1003}{2003}\) <\(\dfrac{-1002}{2003}\) <\(\dfrac{1001}{2002}\) <\(\dfrac{-1003}{-2002}\)