Rút gọn biểu thức:B

Hướng dẫn giải

Ta có: \({2^2} - 1 = 1.3;{\rm{ }}{4^2} - 1 = 3.5;{\rm{ }}{6^2} - 1 = 5.7;....;{\rm{ }}{\left( {n + 1} \right)^2} - 1 = n.\left( {n + 2} \right)\).

Do đó, \(B = \frac{{{1^2}}}{{{2^2} - 1}}.\frac{{{3^2}}}{{{4^2} - 1}}.\frac{{{5^2}}}{{{6^2} - 1}}....\frac{{{n^2}}}{{{{\left( {n + 1} \right)}^2} - 1}} = \frac{{{1^2}}}{{1.3}}.\frac{{{3^2}}}{{3.5}}.\frac{{{5^2}}}{{5.7}}.....\frac{{{n^2}}}{{n\left( {n + 2} \right)}}\)

           \(B = \frac{1}{3}.\frac{3}{5}.\frac{5}{7}....\frac{n}{{n + 2}} = \frac{1}{{n + 2}}.\)

Vậy \(B = \frac{1}{{n + 2}}.\)