\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2002^2}+\dfrac{1}{2003^2}\)

\(A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2001.2002}+\dfrac{1}{2002.2003}\)

\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2001}-\dfrac{1}{2002}+\dfrac{1}{2002}-\dfrac{1}{2003}\)

\(A< 1-\dfrac{1}{2003}< 1\)

Vậy \(A< 1\)

Đặt \(S=\frac{1}{2^2}+\frac{1}{3^2}+......+\frac{1}{2003^2}< \frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{2002.2003}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{2002}-\frac{1}{2003}\)

\(=1-\frac{1}{2003}< 1\)

Vậy S<1

bạn có thể giải rõ ra được ko

ta co 1/2^2<1/1*2+1/3^2+1/2*3+...+1/2003^2 1/2002*2003

1/2^2+1/3^2+...+1/2003^2<1/1*2+1/2*3+...+1/2002*2003

1/2^2+1/3^2+...+1/2003^2<1/1-1/2+1/2-1/3+...+1/2002-1/2003

1/2^2+1/3^2+...+1/2003^2<1-1/2003

1/2^2+1/3^2+...+1/2003^2<2002/2003<1

Vậy 1/2^2+1/3^2+...+1/2003^2<1

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

4S - S = 4 + 4² + 4³ + 4⁴ +....+ 4²⁰⁰² + 4²⁰⁰³ - 1 - 4 - 4² - 4³ - 4⁴ -......- 4²⁰⁰¹ - 4²⁰⁰²

 3S =  4²⁰⁰³ - 1 => 4²⁰⁰³ - 3S = 1  là số nguyên dương nhỏ nhất (đpcm)

làm ơn tách ra giùm mk

=> A<1/1.2 + 1/2.3 + ....+ 1/2001.2002 + 1/2002.2003

=> A< 1- 1/2 + 1/2 - 1/3 + .... + 1/2001 - 1/2002 + 1/2002 - 1/2003

=>A< 1 - 1/2003 < 1

=> A< 1

nguyên một hàng mk đọc ko hỉu????????????

không hiểu......>><

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