Cho $\left( {{u}_{n}} \right)$ là cấp số nhân, đặt...

Gọi $q$ là công bội của cấp số nhân $\left( {{u}_{n}} \right)$.
Ta có ${{S}_{3}}=13\ne 0$ nên ${{u}_{1}}\ne 0$.
Mặt khác
$\left\{ \begin{aligned}
& {{u}_{2}}+{{S}_{4}}=43 \\
& {{S}_{3}}=13 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{u}_{2}}+{{u}_{1}}+{{u}_{2}}+{{u}_{3}}+{{u}_{4}}=43 \\
& {{u}_{1}}+{{u}_{2}}+{{u}_{3}}=13 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& {{u}_{1}}q+{{u}_{1}}+{{u}_{1}}q+{{u}_{1}}{{q}^{2}}+{{u}_{1}}{{q}^{3}}=43 \\
& {{u}_{1}}+{{u}_{1}}q+{{u}_{1}}{{q}^{2}}=13 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& 13{{u}_{1}}\left( 1+2q+{{q}^{2}}+{{q}^{3}} \right)=43{{u}_{1}}\left( 1+q+{{q}^{2}} \right) \\
& {{u}_{1}}+{{u}_{1}}q+{{u}_{1}}{{q}^{2}}=13 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& 13{{q}^{3}}-30{{q}^{2}}-17q-30=0 \\
& {{u}_{1}}+{{u}_{1}}q+{{u}_{1}}{{q}^{2}}=13 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{matrix}
q=3 \\
{{u}_{1}}=1 \\
\end{matrix} \right.$.
Vậy ${{S}_{6}}=\dfrac{{{u}_{1}}\left( 1-{{q}^{6}} \right)}{1-q}=\dfrac{1\left( 1-{{3}^{6}} \right)}{1-3}=364$.