Lời giải:

$A=1+\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{50^2}$
$< 1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}$

$=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}$
=2-\frac{1}{50}< 2$ 

(đpcm)

bạn ơi xem lại bìa đó đi bn ạ

1/2² < 1/2.3 ; 1/3² < 1/3.4 ; .....; 1/50² < 1/50.51  => A < 1+1-1/2+1/2-1/3+...1/50-1/51 < 2  

tổng đài tư vấn có bằng chứng ko 

ko có thì đừng nói

Gọi \(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\) là \(S\)

\(S=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\\ S< \dfrac{1}{1}+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\\ S< 1+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ S< 1+1-\dfrac{1}{50}\\ S< 2-\dfrac{1}{50}< 2\)

Vậy \(S< 2\)

Lời giải:

Đặt \(T=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\)

Dễ thấy:

\(\dfrac{1}{1^2}=1\)

\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)

\(....\)

\(\dfrac{1}{50^2}=\dfrac{1}{50.50}< \dfrac{1}{49.50}\)

\(\Rightarrow T< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)

\(\Rightarrow T< 1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(\Rightarrow T< 1+1-\dfrac{1}{50}\)

\(\Rightarrow T< 2-\dfrac{1}{50}\)

\(\Rightarrow T< 2\)

P=1+1/2+1/3+1/4+...+1/2^100-1

suy ra P=1+1/2+1/3+1/2^2+...+1/2^100+1/2^100-1+1/2^100-1/2^100

suy ra P=1+1/2+(1/3+1/2^2)+(1/5+1/2^3)+...+(1/2^99+1+...+1/2^100)-1/2^100

suy ra P>1+1/2+1/2^2.2+1/2^3.3^2+...+1/2^100.2^99-1/2^100

suy ra P>1+1/2.100-1/2^100

suy ra P>51-1/2^100>51-1

suy ra P>50(đpcm)

Cảm ơn bạn

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