\begin{enumerate}
\item Si $u_0=1$ et $q=2$, calculer $u_0+u_1+\cdots+u_{100}$.

\suite{u}{n}{} est une suite géométrique donc :
\[
u_0+u_1+\cdots+u_n=u_0\times\dfrac{1-q^{n+1}}{1-q}.
\]

Donc ici,
\begin{align*}
u_0+u_1+\cdots+u_{100} & = \dfrac{1-2^{101}}{1-2}\\
\Aboxed{u_0+u_1+\cdots+u_{100} & = 2^{101}-1}
\end{align*}

\item Si $u_0=3$ et $q=\dfrac{1}{2}$, calculer $u_0+u_1+\cdots+u_{50}$.

\begin{align*}
u_0+u_1+\cdots+u_{50} & = 3\times\dfrac{1-\frac{1}{2^{51}}}{1-\frac{1}{2}}\\
& = 3\left(1-\dfrac{1}{2^{51}}\right)\times2\\
\Aboxed{u_0+u_1+\cdots+u_{50} & = 6\left(1-\dfrac{1}{2^{51}}\right)}
\end{align*}
\newpage
\item Si $u_1=60$ et $q=\dfrac{1}{3}$, calculer $u_1+u_2+\cdots+u_{100}$.

\begin{align*}
u_1+u_2+\cdots+u_{100} &= u_1\times\dfrac{1-q^{100}}{1-q}\\
& = 60\times\dfrac{1-\frac{1}{3^{100}}}{1-\frac{1}{3}}\\
& = 60\left(1-\dfrac{1}{3^{100}}\right)\times\dfrac{3}{2}\\
\Aboxed{u_1+u_2+\cdots+u_{100} & = 45\left(1-\dfrac{1}{3^{100}}\right)}
\end{align*}

\item Si $u_1=50$ et $q=10$, calculer $u_1+u_2+\cdots+u_{50}$.

\begin{align*}
u_1+u_2+\cdots+u_{50} & = 50\times\dfrac{1-10^{50}}{1-10}\\
\Aboxed{u_1+u_2+\cdots+u_{5} & = \dfrac{50}{9}\big(10^{50}-1\big)}
\end{align*}
\end{enumerate}