Question

[vũ thị minh anh]

cho A=1/101+1/102+1/103+...+1/200

Chứng minh rằng:

a)A>7/12

b)A>5/8

Answer 1

[Phạm Hương Giang]

Ta có: $C=\genfrac{}{}{0ex}{}{1}{101}+\genfrac{}{}{0ex}{}{1}{102}+\genfrac{}{}{0ex}{}{1}{103}+...+\genfrac{}{}{0ex}{}{1}{200}$

$=\left(\genfrac{}{}{0ex}{}{1}{101}+\genfrac{}{}{0ex}{}{1}{102}+...+\genfrac{}{}{0ex}{}{1}{120}\right)+\left(\genfrac{}{}{0ex}{}{1}{121}+\genfrac{}{}{0ex}{}{1}{122}+\genfrac{}{}{0ex}{}{1}{123}+...+\genfrac{}{}{0ex}{}{1}{150}\right)+\left(\genfrac{}{}{0ex}{}{1}{151}+\genfrac{}{}{0ex}{}{1}{152}+\genfrac{}{}{0ex}{}{1}{153}+...+\genfrac{}{}{0ex}{}{1}{180}\right)+\left(\genfrac{}{}{0ex}{}{1}{181}+\genfrac{}{}{0ex}{}{1}{182}+\genfrac{}{}{0ex}{}{1}{183}+...+\genfrac{}{}{0ex}{}{1}{200}\right)$

$\iff C>20\cdot \genfrac{}{}{0ex}{}{1}{120}+30\cdot \genfrac{}{}{0ex}{}{1}{150}+30\cdot \genfrac{}{}{0ex}{}{1}{180}+20\cdot \genfrac{}{}{0ex}{}{1}{200}$

$\iff C>\genfrac{}{}{0ex}{}{1}{6}+\genfrac{}{}{0ex}{}{1}{5}+\genfrac{}{}{0ex}{}{1}{6}+\genfrac{}{}{0ex}{}{1}{10}=\genfrac{}{}{0ex}{}{19}{30}=\genfrac{}{}{0ex}{}{76}{120}$

$\iff C>\genfrac{}{}{0ex}{}{75}{120}=\genfrac{}{}{0ex}{}{5}{8}$

hay $C>\genfrac{}{}{0ex}{}{5}{8}$(đpcm)

 TỰ thay C=a nhA