Question

Chứng minh rằng : 

\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}...+\frac{1}{100^2}<\frac{1}{2}\)

Mong các bạn giúp mình! Cám ơn các bạn nhiều !!!

Answer 1

\(\frac{1}{3^2}<\frac{1}{3.4}\)

\(\frac{1}{4^2}<\frac{1}{4.5}\)

\(\frac{1}{5^2}<\frac{1}{5.6}\)

\(...\)

\(\frac{1}{100^2}<\frac{1}{100.101}\)

\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{100.101}\)

\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{100}-\frac{1}{101}\)

\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{3}-\frac{1}{101}\)

Mà \(\frac{1}{3}<\frac{1}{2}\) nên \(\frac{1}{3}-\frac{1}{101}<\frac{1}{2}\)

hay \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{2}\)

Answer 2

Đặt A=1/3^2+1/4^2+1/5^2+...+1/100^2

Suy raA<1/2*3+1/3*4+1/4*5+..+1/99*100

A<1/2-1/100<1/2

Ta có điều phải chứng minh.