Question

1. Tính

K = \(\frac{1}{7}\)+\(\frac{1}{91}\)+\(\frac{1}{247}\)+\(\frac{1}{475}\)+\(\frac{1}{755}\)+\(\frac{1}{1147}\)

2. Chứng minh rằng

\(\frac{1}{4}\)+\(\frac{1}{16}\)+\(\frac{1}{36}\)+\(\frac{1}{64}\)+\(\frac{1}{100}\)+\(\frac{1}{144}\)+\(\frac{1}{196}\)<\(\frac{1}{2}\)

 

 

 

 

 

 

 

 

 

 

 

Answer 1

Ta có:

  \(K=\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{775}+\frac{1}{1147}\)

     \(=\frac{1}{1.7}+\frac{1}{7.13}+\frac{1}{13.19}+\frac{1}{19.25}+\frac{1}{25.31}+\frac{1}{31.37}\)

     \(=\frac{1}{6}\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+\frac{1}{19}-\frac{1}{25}+\frac{1}{25}-\frac{1}{31}+\frac{1}{31}-\frac{1}{37}\right)\)

      \(=\frac{1}{6}.\left(1-\frac{1}{37}\right)=\frac{1}{6}.\frac{36}{37}=\frac{6}{37}\)

Answer 2

K = \(\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{755}+\frac{1}{1147}=0,1621963429\)