Tính \(I = \lim \left( {\dfrac{1}{{{n^2}}} + \dfrac{3}{{{n^2}}} + \dfrac{5}{{{n^2}}} + ... + \dfrac{{2n +
Tính \(I = \lim \left( {\dfrac{1}{{{n^2}}} + \dfrac{3}{{{n^2}}} + \dfrac{5}{{{n^2}}} + ... + \dfrac{{2n + 1}}{{{n^2}}}} \right)\).
Đáp án đúng là:
\(\begin{array}{l}I = \lim \left( {\dfrac{1}{{{n^2}}} + \dfrac{3}{{{n^2}}} + \dfrac{5}{{{n^2}}} + ... + \dfrac{{2n + 1}}{{{n^2}}}} \right)\\I = \lim \dfrac{{1 + 3 + 5 + ... + \left( {2n + 1} \right)}}{{{n^2}}}\\I = \lim \dfrac{{\left( {1 + 2n + 1} \right).n}}{{2{n^2}}}\\I = \lim \dfrac{{1 + n}}{n} = \lim \dfrac{{\dfrac{1}{n} + 1}}{1} = 1\end{array}\)
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