Question

\(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{399}{400}\)  Chứng tỏ tổng trên >18 

Answer 1

Xét A= \(\frac{3}{4}\)+ \(\frac{8}{9}\) +...+ \(\frac{399}{400}\)

= (1 - \(\frac{1}{2^2}\)) + (1- \(\frac{1}{3^2}\)) +...+ (1- \(\frac{1}{20^2}\))

= (1+1+1+...+1) - (\(\frac{1}{2^2}\) + \(\frac{1}{3^2}\)+...+ \(\frac{1}{20^2}\)) Bạn phải mở ngoặc có 19 số 1 nha!

= 19 - (\(\frac{1}{2^2}\) + \(\frac{1}{3^2}\)+...+ \(\frac{1}{20^2}\))  

Đặt B =\(\frac{1}{2^2}\) + \(\frac{1}{3^2}\)+...+ \(\frac{1}{20^2}\) < \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) +...+ \(\frac{1}{19.20}\) = 1- \(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{3}\) +...+ \(\frac{1}{19}\) - \(\frac{1}{20}\) = 1 - \(\frac{1}{20}\) = \(\frac{19}{20}\)

=> A= 19 - B= 18+ 1- \(\frac{19}{20}\) >18 => A>18