Giải thích các bước giải:
Tổng của CSN lùi vô hạn được tính bởi công thức: \({S_n} = {u_1}.\frac{1}{{1 - q}}\)
\(\begin{array}{l}
86,\\
\left\{ \begin{array}{l}
{S_n} = 2\\
{u_1} + {u_2} + {u_3} = \frac{9}{4}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{u_1}.\frac{1}{{1 - q}} = 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\left| q \right| < 1} \right)\\
{u_1}\left( {1 + q + {q^2}} \right) = \frac{9}{4}
\end{array} \right.\\
\Rightarrow \left( {1 + q + {q^2}} \right):\frac{1}{{1 - q}} = \frac{9}{4}:2\\
\Leftrightarrow \left( {1 + q + {q^2}} \right)\left( {1 - q} \right) = \frac{9}{8}\\
\Leftrightarrow 1 - {q^3} = \frac{9}{8}\\
\Leftrightarrow {q^3} = - \frac{1}{8}\\
\Leftrightarrow q = - \frac{1}{2} \Rightarrow {u_1} = 3\\
87,\\
S = 9 + 3 + 1 + \frac{1}{3} + \frac{1}{9} + .... + \frac{1}{{{3^{n - 3}}}}\\
{u_1} = 9;\,\,\,\,\,q = \frac{1}{3} < 1\\
\Rightarrow S = \frac{{{u_1}}}{{1 - q}} = \frac{9}{{1 - \frac{1}{3}}} = \frac{9}{{\frac{2}{3}}} = \frac{{27}}{2}\\
88,\\
S = \frac{1}{{\sqrt 2 }}\left( {1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ..... + \frac{1}{{{2^n}}}} \right)\\
{S_1} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + .... + \frac{1}{{{2^n}}}\\
{u_1} = 1;\,\,\,\,q = \frac{1}{2} < 1\\
\Rightarrow {S_n} = \frac{{{u_1}}}{{1 - q}} = \frac{1}{{1 - \frac{1}{2}}} = 2\\
\Rightarrow S = \sqrt 2 .{S_n} = 2\sqrt 2 \\
89,\\
S = 1 + \frac{2}{3} + \frac{4}{9} + ..... + \frac{{{2^n}}}{{{3^n}}} + ....\\
= 1 + \frac{2}{3} + {\left( {\frac{2}{3}} \right)^2} + ..... + {\left( {\frac{2}{3}} \right)^n} + ....\\
{u_1} = 1;\,\,\,q = \frac{2}{3} < 0\\
\Rightarrow S = \frac{{{u_1}}}{{1 - q}} = \frac{1}{{1 - \frac{2}{3}}} = 3
\end{array}\)
