
\begin{enumerate}
\item $u_1=\dfrac{2}{7}u_0+6=\dfrac{2}{7}\times2+6=\dfrac{4}{7}+\dfrac{42}{7}=\dfrac{46}{7}$.

$u_2=\dfrac{2}{7}u_1+6=\dfrac{2}{7}\times\dfrac{46}{7}+6=\dfrac{92}{49}+\dfrac{294}{49}=\dfrac{386}{49}$.

\item $v_n=u_n-\dfrac{42}{5}$.
\begin{enumerate}
\item $\begin{aligned}[t]
v_{n+1} & = u_{n+1}-\dfrac{42}{5}\\
& = \dfrac{2}{7}u_n+6-\dfrac{42}{5}\\
& = \dfrac{2}{7}u_n+\dfrac{30}{5}-\dfrac{42}{5}\\
&=  \dfrac{2}{7}u_n-\dfrac{12}{5}\\
& = \dfrac{2}{7}\left(u_n-\dfrac{12}{5}\times\dfrac{7}{2}\right)\\
&=  \dfrac{2}{7}{\color{red}\left(u_n-\dfrac{42}{5}\right)} = \dfrac{2}{7}{\color{red}v_n}.
\end{aligned}$
\newpage
Ainsi, \suite{v}{n}{} est une suite géométrique de raison $q=\dfrac{2}{7}$ et de premier terme :\linebreak $v_0=u_0-\dfrac{42}{5}=2-\dfrac{42}{5}=-\dfrac{32}{5}$.

\item On déduit que $v_n=v_0\times q^n$, soit $\boxed{v_n=-\dfrac{32}{5}\times\left(\dfrac{2}{7}\right)^n}$.

Or, $v_n=u_n-\dfrac{42}{5}$ donc $u_n=v_n+\dfrac{42}{5}$; ainsi, $\boxed{u_n=\dfrac{42}{5}-\dfrac{32}{5}\times\left(\dfrac{2}{7}\right)^n}$.
\end{enumerate}

\item
\begin{enumerate}
\item $S$ est la somme des premiers termes d'une suite géométrique donc :
\begin{align*}
S & =v_0\times\dfrac{1-q^{11}}{1-q}\\
& = -\dfrac{32}{5}\times\dfrac{1-\left(\frac{2}{7}\right)^{11}}{1-\frac{2}{7}}
\end{align*}

\begin{align*}
S & = -\dfrac{32}{5}\times\dfrac{1-\left(\frac{2}{7}\right)^{11}}{\frac{5}{7}}\\
& = -\dfrac{32}{5}\times\dfrac{7}{5}\times\left[1-\left(\frac{2}{7}\right)^{11}\right]\\
\Aboxed{S & = -\dfrac{134}{25}\left[1-\left(\frac{2}{7}\right)^{11}\right]}
\end{align*}

\item On a :
\begin{align*}
S' & = \displaystyle\sum_{k=0}^{10}\left(v_k-\dfrac{42}{5}\right)\qquad\text{car }u_k=v_k-\dfrac{42}{5}\\
   & = \displaystyle\sum_{k=0}^{10} v_k - \displaystyle\sum_{k=0}^{10}\dfrac{42}{5}\qquad\text{(on a découpé la somme en deux)}\\
   & = S'-\dfrac{42}{5}\times11\qquad\text{ car la dernière somme signifie que l'on ajoute}\\
   & \hspace*{7.5cm}\text{ 11 fois la fraction }\dfrac{42}{5}\\
   & =  -\dfrac{134}{25}\left[1-\left(\frac{2}{7}\right)^{11}\right]-\dfrac{462}{5}\\
   & = -\dfrac{134}{25} + \dfrac{134}{25}\times\left(\frac{2}{7}\right)^{11}-\dfrac{462}{5}\\
   & = \dfrac{134}{25}\times\left(\frac{2}{7}\right)^{11}-\dfrac{134}{25}-\dfrac{462\times5}{5\times5}\\
\Aboxed{ S'   & = \dfrac{134}{25}\times\left(\frac{2}{7}\right)^{11}-\dfrac{2444}{25}}
\end{align*}
\end{enumerate}
\end{enumerate}
