Tính \(B = \dfrac{1}{{A_2^2}} + \dfrac{1}{{A_3^2}} + ... + \dfrac{1}{{A_n^2}}\), biết \(C_n^1 + 2\dfrac{{C_n^2}}{{C_n^1}} + ... + n\dfrac{{C_n^n}}{{C_n^{n - 1}}} = 55\)

Ta có: \(C_n^1 = n\); \(2\dfrac{{C_n^2}}{{C_n^1}} = 2.\dfrac{{\dfrac{{n!}}{{2!.(n - 2)!}}}}{{\dfrac{{n!}}{{1!.(n - 1)!}}}} = n - 1\);...; \(n\dfrac{{C_n^n}}{{C_n^{n - 1}}} = n.\dfrac{1}{{\dfrac{{n!}}{{1!.(n - 1)!}}}} = 1\)

Nên \(C_n^1 + 2\dfrac{{C_n^2}}{{C_n^1}} + ... + n\dfrac{{C_n^n}}{{C_n^{n - 1}}} = 55\) \( \Leftrightarrow n + \left( {n - 1} \right) + \left( {n - 2} \right) + ... + 2 + 1 = 55\) \( \Leftrightarrow \dfrac{{n\left( {n + 1} \right)}}{2} = 55 \Leftrightarrow n = 10\)

Lại có: \(A_2^2 = 2! = 2.1\); \(A_3^2 = \dfrac{{3!}}{{\left( {3 - 2} \right)!}} = 3.2\); \(A_4^2 = \dfrac{{4!}}{{\left( {4 - 2} \right)!}} = 4.3\);…;\(A_{10}^2 = \dfrac{{10!}}{{\left( {10 - 2} \right)!}} = 10.9\)

\( \Rightarrow B = \dfrac{1}{{A_2^2}} + \dfrac{1}{{A_3^2}} + ... + \dfrac{1}{{A_{10}^2}}\) \(=\dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + ... + \dfrac{1}{{9.10}}\) \( = 1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{9} - \dfrac{1}{{10}} = 1 - \dfrac{1}{{10}} = \dfrac{9}{{10}}\)