Cho số nguyên dương n thỏa mãn \(C_n^0 + \frac{{C_n^1}}{{1 + 1}} + \frac{{C_n^2}}{{1 + 2}} + \ldots + \frac{{C_n^n}}{{1 + n}} = \frac{{{2^{100}} - 1}}{{100}}\). Giá trị của \[n\] bằng bao nhiêu?

Đáp án: ……….

Ta có \(\int\limits_0^1 {{{\left( {1 + x} \right)}^n}dx} = \int\limits_0^1 {{{\left( {1 + x} \right)}^n}d\left( {1 + x} \right)} = \left. {\frac{{{{\left( {1 + x} \right)}^{n + 1}}}}{{n + 1}}} \right|_0^1 = \frac{{{2^{n + 1}} - 1}}{{n + 1}}\)

Mặt khác, \({\left( {1 + x} \right)^n} = C_n^0 + C_n^1 \cdot x + C_n^2 \cdot {x^2} + \ldots + C_n^n \cdot {x^n}\)

\[ \Leftrightarrow \int\limits_0^1 {{{\left( {1 + x} \right)}^n}dx} = C_n^0\int\limits_0^1 {dx} + C_n^1\int\limits_0^1 {x \cdot dx} + C_n^2\int\limits_0^1 {{x^2}dx} + \ldots + C_n^n\int\limits_0^1 {{x^n}dx} \]

\( \Leftrightarrow \int\limits_0^1 {{{\left( {1 + x} \right)}^n}dx} = \left. {C_n^0x} \right|_0^1 + \left. {\frac{{C_n^1{x^2}}}{2}} \right|_0^1 + \left. {\frac{{C_n^2{x^3}}}{3}} \right|_0^1 + \ldots + \left. {\frac{{C_n^n{x^{n + 1}}}}{{n + 1}}} \right|_0^1\)

\( \Leftrightarrow \frac{{{2^{n + 1}} - 1}}{{n + 1}} = C_n^0 + \frac{{C_n^1}}{{1 + 1}} + \frac{{C_n^2}}{{1 + 2}} + \ldots + \frac{{C_n^n}}{{n + 1}} \Leftrightarrow \frac{{{2^{n + 1}} - 1}}{{n + 1}} = \frac{{{2^{100}} - 1}}{{100}} \Leftrightarrow n = 99.\)

Đáp án: 99.