Giải thích các bước giải:

a,

\({u_n} = \frac{5}{{{2^n}}} = \frac{5}{2}.\frac{1}{{{2^{n - 1}}}} = \frac{5}{2}.{\left( {\frac{1}{2}} \right)^{n - 1}}\)

Do đó, \(\left( {{u_n}} \right)\) là CSN có \({u_1} = \frac{5}{2};\,\,\,q = \frac{1}{2}\)

b,

\(\begin{array}{l}
\left\{ \begin{array}{l}
{u_3} + {u_5} = 90\\
{u_2} - {u_6} = 240
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{u_1}{q^2} + {u_q}{q^4} = 90\\
{u_1}q - {u_1}{q^5} = 240
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{u_1}{q^2}\left( {1 + {q^2}} \right) = 90\\
{u_1}q\left( {1 - {q^4}} \right) = 240
\end{array} \right.\\
 \Rightarrow \frac{{{u_1}{q^2}\left( {1 + {q^2}} \right)}}{{{u_1}q\left( {1 - {q^4}} \right)}} = \frac{{90}}{{240}}\\
 \Leftrightarrow \frac{{q\left( {1 + {q^2}} \right)}}{{\left( {1 - {q^2}} \right)\left( {1 + {q^2}} \right)}} = \frac{3}{8}\\
 \Leftrightarrow \frac{q}{{1 - {q^2}}} = \frac{3}{8}\\
 \Leftrightarrow  - 3{q^2} - 8q + 3 = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
q =  - 3 \Rightarrow {u_1} = 1 \Rightarrow {u_{15}} = {u_1}.{q^{14}} = {3^{14}}\\
q = \frac{1}{3} \Rightarrow {u_1} = 729 \Rightarrow {u_{15}} = {u_1}.{q^{14}} = \frac{1}{{{3^8}}}
\end{array} \right.\\
{S_{20}} = {u_1} + {u_2} + {u_3} + ..... + {u_{20}}\\
 = {u_1} + {u_1}q + {u_1}{q^2} + ..... + {u_1}{q^{19}}\\
 = {u_1}.\left( {1 + q + {q^2} + ..... + {q^{19}}} \right)\\
 = {u_1}.\frac{{{q^{20}} - 1}}{{q - 1}}\\
q =  - 3 \Rightarrow {S_{20}} = \frac{{{3^{20}} - 1}}{{ - 4}} = \frac{{1 - {3^{20}}}}{4}\\
q = \frac{1}{3} \Rightarrow {S_{20}} = 729.\frac{{{{\left( {\frac{1}{3}} \right)}^{20}} - 1}}{{\left( {\frac{1}{3}} \right) - 1}} = \frac{{2187.\left( {1 - \frac{1}{{{3^{20}}}}} \right)}}{2}
\end{array}\)