Bài tập 3.29 trang 131 SBT Toán 11

a) \(\left\{ \begin{array}{l}
{u_n} = {u_1}{2^{n - 1}} = 96\\
{S_n} = \frac{{{u_1}\left( {{2^n} - 1} \right)}}{{2 - 1}} = 189
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{u_1}{.2^n} = 192\\
{u_1}{.2^n} - {u_1} = 189
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{u_1} = 3\\
{2^n} = 64
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{u_1} = 3\\
n = 6
\end{array} \right.\)

b) Ta có:

\(\begin{array}{l}
\left\{ \begin{array}{l}
{u_n} = {u_1}{q^{n - 1}} = \frac{1}{8}\\
{S_n} = \frac{{{u_1}\left( {{q^n} - 1} \right)}}{{q - 1}} = \frac{{31}}{8}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2.{q^{n - 1}} = \frac{1}{8}\\
\frac{{2\left( {{q^n} - 1} \right)}}{{q - 1}} = \frac{{31}}{8}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{q^{n - 1}} = \frac{1}{{16}}\\
2{q^n} - 2 = \frac{{31}}{8}\left( {q - 1} \right)
\end{array} \right.\\
 \Leftrightarrow 2.\frac{1}{{16}}.q - 2 = \frac{{31}}{8}\left( {q - 1} \right) \Leftrightarrow q = \frac{1}{2}\\
 \Rightarrow {\left( {\frac{1}{2}} \right)^{n - 1}} = \frac{1}{{16}} \Leftrightarrow n - 1 = 4 \Leftrightarrow n = 5
\end{array}\)

-- Mod Toán 11 HỌC247