Question

[đức hiếu đỗ]

cần gấp giúp mik nha

Answer 1

[Đỗ Tuệ Lâm]

a) . 

Ta có :

\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+....+\dfrac{1}{200^2}< \dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{199.200}\)

Gọi \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{199.200}\left(là.B\right)\)

 ta có:

\(B=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{199}-\dfrac{1}{200}\)

\(\dfrac{1}{2}-\dfrac{1}{200}< \dfrac{1}{2}\) (1 ) 

Từ (1) ta suy ra được :

\(A=\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+....+\dfrac{1}{200^2}< \dfrac{1}{2}\left(đ.p.c.m\right)\)

Answer 2

[Đỗ Tuệ Lâm]

b ) .

Ta có :

\(B=2\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{200^2}\right)\)\(>2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{199.200}\right)\)

gọi vế thứ (2) ở B là C 

Ta có :

\(C=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{199}-\dfrac{1}{200}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{200}\right)=\dfrac{99}{100}>\dfrac{9}{20}\) (1)

Từ (1) ta suy ra được :

\(B=\dfrac{2}{2^2}+\dfrac{2}{3^2}+\dfrac{2}{4^2}+.....+\dfrac{2}{200^2}>\dfrac{9}{20}\)