a) Ta có: \({u_n} = {u_1}{q^{n - 1}} \Leftrightarrow 96 = {u_1}{.2^{n - 1}}\)
Lại có: \({S_n} = \dfrac{{{u_1}\left( {{q^n} - 1} \right)}}{{q - 1}}\) \( \Leftrightarrow 189 = \dfrac{{{u_1}\left( {{2^n} - 1} \right)}}{{2 - 1}}\) \( \Leftrightarrow 189 = {u_1}\left( {{2^n} - 1} \right)\)
\( \Rightarrow \dfrac{{189}}{{96}} = \dfrac{{{2^n} - 1}}{{{2^{n - 1}}}}\) \( \Leftrightarrow {189.2^{n - 1}} = {96.2^{n - 1}}.2 - 96\) \( \Leftrightarrow {3.2^{n - 1}} = 96 \Leftrightarrow {2^{n - 1}} = 32\) \( \Leftrightarrow n - 1 = 5 \Leftrightarrow n = 6\)
Vậy \(n = 6.\)
b) Ta có hệ: \(\left\{ \begin{array}{l}\dfrac{1}{8} = 2.{q^{n - 1}}\\\dfrac{{31}}{8} = \dfrac{{2\left( {{q^n} - 1} \right)}}{{q - 1}}\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}{q^{n - 1}} = \dfrac{1}{{16}}\\31\left( {q - 1} \right) = 16\left( {q.{q^{n - 1}} - 1} \right)\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}{q^{n - 1}} = \dfrac{1}{{16}}\\31\left( {q - 1} \right) = 16\left( {\dfrac{1}{{16}}q - 1} \right)\end{array} \right.\)
\( \Leftrightarrow \left\{ \begin{array}{l}{q^{n - 1}} = \dfrac{1}{{16}}\\30q = 15\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}q = \dfrac{1}{2}\\n - 1 = 4\end{array} \right. \Rightarrow n = 5\)
Vậy \(n = 5.\)
