a) Ta có : \({S_n} = \dfrac{1}{2} + \dfrac{3}{{{2^2}}} + \dfrac{5}{{{2^3}}} + ... + \dfrac{{2n - 1}}{{{2^n}}}\)
\(2{S_n} = 1 + \dfrac{3}{2} + \dfrac{5}{{{2^2}}} + ... + \dfrac{{2n - 1}}{{{2^{n - 1}}}}\)
\( \Rightarrow 2{S_n} - {S_n}\) \( = 1 + 1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + ... + \dfrac{1}{{{2^{n - 2}}}} - \dfrac{{2n - 1}}{{{2^n}}}\)
\( \Rightarrow {S_n} = 2 + \dfrac{{\dfrac{1}{2}\left( {{{\left( {\dfrac{1}{2}} \right)}^{n - 2}} - 1} \right)}}{{\dfrac{1}{2} - 1}} - \dfrac{{2n - 1}}{{{2^n}}}\) \( = 2 - {\left( {\dfrac{1}{2}} \right)^{n - 2}} + 1 - \dfrac{{2n - 1}}{{{2^n}}}\) \( = 3 - \dfrac{{2n + 3}}{{{2^n}}}\)
b) Nếu \(n = 2k + 1\) thì :
\({S_n} = {1^2} - {2^2} + {3^2} - {4^2} + ... + {\left( {2k + 1} \right)^2}\)
\( = - 3 - 7 - 11 - ... - \left( {4k - 1} \right) + {\left( {2k + 1} \right)^2}\)
Dãy tổng \( - 3 - 7 - 11 - ... - \left( {4k - 1} \right)\) là dãy tổng \(k\) số hạng đầu của cấp số cộng có \({u_1} = - 3,d = - 4\) nên \( - 3 - 7 - 11 - ... - \left( {4k - 1} \right)\) \( = \dfrac{{k\left[ {2.\left( { - 3} \right) + \left( {k - 1} \right).\left( { - 4} \right)} \right]}}{2} = k\left( { - 2k - 1} \right)\)
Do đó \({S_n} = k\left( { - 2k - 1} \right) + {\left( {2k + 1} \right)^2}\) \( = 2{k^2} + 3k + 1\)
Nếu \(n = 2k\) thì \({S_n} = {1^2} - {2^2} + {3^2} - {4^2} + ... + {\left( {2k - 1} \right)^2} - {\left( {2k} \right)^2}\)
\( = - 3 - 7 - 11 - ... - \left( {4k - 1} \right)\) \( = k\left( { - 2k - 1} \right) = - 2{k^2} - k\)
Vậy \({S_n} = \left\{ \begin{array}{l}2{k^2} + 3k + 1\,neu\,n = 2k + 1\\ - 2{k^2} - k\,neu\,n = 2k\end{array} \right.\)
