Cho \({a_1};{a_2};...;{a_{2024}}\) là 2024 số thực thỏa mãn \({a_k} = \frac{{2k + 1}}{{{{\left( {{k^2} + k} \right)}^2}}}\) với \(k \in \left\{ {1;2;...;2024} \right\}\).
Tính tổng \({S_{2024}} = {a_1} + {a_2} + {a_3} + ... + {a_{2024}}\).
Phân tích \({a_k} = \frac{{2k + 1}}{{{{\left( {{k^2} + k} \right)}^2}}} = \frac{1}{{{k^2}}} - \frac{1}{{{{\left( {k + 1} \right)}^2}}}\)
Từ đó tính \({S_{2024}}\).
Ta có:
\({a_k} = \frac{{2k + 1}}{{{{\left( {{k^2} + k} \right)}^2}}} = \frac{{2k + 1}}{{{{\left[ {k\left( {k + 1} \right)} \right]}^2}}} = \frac{{{{\left( {k + 1} \right)}^2} - {k^2}}}{{{k^2}{{\left( {k + 1} \right)}^2}}} = \frac{1}{{{k^2}}} - \frac{1}{{{{\left( {k + 1} \right)}^2}}}\)
Do đó:
\(\begin{array}{l}{S_{2024}} = {a_1} + {a_2} + {a_3} + ... + {a_{2024}}\\ = \left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) + \left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right) + \left( {\frac{1}{{{3^2}}} - \frac{1}{{{4^2}}}} \right) + ... + \left( {\frac{1}{{{{2023}^2}}} - \frac{1}{{{{2024}^2}}}} \right)\\ = 1 - \frac{1}{{{{2024}^2}}}\\ = \frac{{{{2024}^2} - 1}}{{{{2024}^2}}}\end{array}\)
Vậy \({S_{2024}} = \frac{{{{2024}^2} - 1}}{{{{2024}^2}}}\)
