Question

Chứng minh rằng:

\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+...+\(\dfrac{1}{100^2}\)<1

Answer 1

Ta có: 1/2² < 1/ 1.2

1/3² < 1/2.3

1/4² < 1/3.4

....

1/ 100² < 1/ 99.100

Nên A< 1/1.2+1/2.3+...+1/99.100

= 1- 1/2+1/2 -1/3+1/3 -1/4+...+1/99-1/100

= 1- 1/100

<1 Vậy A><1. >

Ma 1 > 1/100

Vay…

Answer 2

\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}221​+321​+421​+...+10021​<1.21​+2.31​+3.41​+...+99.1001​
=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}< 1=1−21​+21​−31​+31​−41​+...+991​−1001​=1−1001​<1

Answer 3

\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}221​+321​+421​+...+10021​<1.21​+2.31​+3.41​+...+99.1001​
=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}< 1=1−21​+21​−31​+31​−41​+...+991​−1001​=1−1001​<1

Answer 4

Đặt biểu thức là \(A\)

Ta có:\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)

\(A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{100.100}\)

\(A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

\(A< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A< \dfrac{1}{1}-\dfrac{1}{100}\)

\(A< \dfrac{100}{100}-\dfrac{1}{100}< 1\)

\(A< \dfrac{99}{100}< 1\)

\(\Rightarrow A< 1\)

\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1\)

Answer 5

Ta có: 1/22 < 1/ 1.2

1/32 < 1/2.3

1/42 < 1/3.4

....

1/ 1002 < 1/ 99.100

Nên A< 1/1.2+1/2.3+...+1/99.100

= 1- 1/2+1/2 -1/3+1/3 -1/4+...+1/99-1/100

= 1- 1/100

<1 Vậy A><1. >

Ma 1 > 1/100

Vậy…

Answer 6

Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)

\(\Rightarrow A< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)

\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Rightarrow A< 1-\dfrac{1}{100}\)

\(\Rightarrow A< \dfrac{99}{100}< 1\left(đpcm\right)\)

Answer 7

Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)

\(\Rightarrow A< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)

\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Rightarrow A< 1-\dfrac{1}{100}< 1\left(đpcm\right)\)