Ta có: \(\frac{1}{2}+\frac{1}{3}< 2\cdot\frac{1}{2}=1\)
\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}< 4\cdot\frac{1}{4}=1\)
\(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+...+\frac{1}{15}< 8\cdot\frac{1}{8}=1\)
\(\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+...+\frac{1}{31}< 16\cdot\frac{1}{16}=1\)
\(\frac{1}{32}+\frac{1}{33}+\frac{1}{34}+...+\frac{1}{63}< 32\cdot\frac{1}{32}=1\)
Cộng từng vế của các BĐT trên ta có:
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 5\)
\(\Leftrightarrow64+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 69\)
\(\Leftrightarrow1+\frac{1}{1}+1+\frac{1}{2}+1+\frac{1}{3}+...+1+\frac{1}{63}< 69\)
\(\Leftrightarrow\frac{2}{1}+\frac{3}{2}+\frac{4}{3}+...+\frac{64}{63}< 69\)
\(\Leftrightarrow\frac{2^2}{1\cdot2}+\frac{3^2}{2\cdot3}+\frac{4^2}{3\cdot4}+...+\frac{64^2}{63\cdot64}< 69\)đpcm
Cho Linh xin 2 k nào :D
