Question

[Free Fire]

Cho G =1/100^2+1/101^2+1/102^2+....+1/198^2+1/199^2 . CMR 1/200 bé hơn G bé hơn 1/99

Giúp mk với nha.

Answer 1

[.]

Ta có : \(\frac{1}{100^2}< \frac{1}{99.100}\)

            \(\frac{1}{101^2}< \frac{1}{100.101}\)

            \(\frac{1}{102^2}< \frac{1}{101.102}\)

             ...

           \(\frac{1}{198^2}< \frac{1}{197.198}\)

           \(\frac{1}{199^2}< \frac{1}{198.199}\)

\(\Rightarrow G< \frac{1}{99.100}+\frac{1}{100.101}+\frac{1}{101.102}+...+\frac{1}{197.198}+\frac{1}{198.199}\)

\(\Rightarrow G< \frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+...+\frac{1}{198}-\frac{1}{199}\)

\(\Rightarrow G< \frac{1}{99}-\frac{1}{199}< \frac{1}{99}\)(1)

Ta có : \(\frac{1}{100^2}>\frac{1}{100.101}\)

            \(\frac{1}{101^2}>\frac{1}{101.102}\)

            \(\frac{1}{102^2}>\frac{1}{102.103}\)

             ...

            \(\frac{1}{199^2}>\frac{1}{199.200}\)

\(\Rightarrow G>\frac{1}{100.101}+\frac{1}{101.102}+\frac{1}{102.103}+...+\frac{1}{199.200}\)

\(\Rightarrow G>\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+\frac{1}{102}-\frac{1}{103}+...+\frac{1}{199}-\frac{1}{200}\)

\(\Rightarrow G>\frac{1}{100}-\frac{1}{200}=\frac{1}{200}\)(2)

Từ (1) và (2)

\(\Rightarrow\frac{1}{200}< G< \frac{1}{99}\)

Vậy \(\frac{1}{200}< G< \frac{1}{99}\).