Question

Tính giúp tôi vớiCMR:

1/4²+1/5²+1/6²+1/7²+....+1/100²<1/3

 

 

Answer 1

Cho \(Q=\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)

Ta có: \(R=\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)

\(R=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(R=\dfrac{1}{3}-\dfrac{1}{100}\)

\(\Rightarrow R=\dfrac{1}{3}-\dfrac{1}{100}< \dfrac{1}{3}\) (a) 

Và \(Q=\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< R=\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\) (b)

Từ (a) và (b) \(\Rightarrow Q< R< \dfrac{1}{3}\)

Answer 2

1/4^2<1/3*4

1/5^2<1/4*5

...

1/100^2<1/99*100

=>A<1/3-1/4+1/4-1/5+...+1/99-1/100=97/300<1/3