Question

Cho A=\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\) Chứng minh A<2

Answer 1

\(\frac{1}{2^2}< \frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2}-\frac{1}{3}\)
\(.......\)
\(\frac{1}{50^2}< \frac{1}{49}-\frac{1}{50}\)

\(\Rightarrow A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A< \frac{1}{1}-\frac{1}{50}=\frac{49}{50}\)
Mà \(\frac{49}{50}< 2\)
\(\Rightarrow A< 2\)
 

Answer 2

a<2 ai k cho mik, mik se k lại hứa thế lun nói là làm

Answer 3

ta có:1/1^2=1/1

1/2^2=1/2*2<1/1*2=1/1-1/2

1/3^2=1/3*3<1/2*3=1/2-1/3

1/4^2=1/4*4<1/3*4

...

1/50^2=1/50*50<1/49*50=1/49-1/50

=>A=1/1-1/50+1

A=99/50<100/50=2

=>A<2

vậy A<2

Answer 4

Đặt \(B=1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)

Ta có : \(\frac{1}{1^2}=1;\frac{1}{2^2}>\frac{1}{1\cdot2};\frac{1}{3^2}>\frac{1}{3\cdot4}\)

...

\(\frac{1}{50^2}< \frac{1}{49\cdot50}\)

\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)

\(B=1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)

\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{49}-\frac{1}{50}\)

\(=2-\frac{1}{50}< 2\)

\(\Rightarrow A< B< 2\)