[tex]\displaystyle\\
\frac{1}{5}+ \frac{1}{5^2}+ \frac{1}{5^3}+\cdots + \frac{1}{5^{100}}=?\\\\
\text{Este o progresie geometrica cu: }\\\\
\begin{cases}
b_1 = \dfrac{1}{5}\\\\
q = \dfrac{1}{5}\\\\
\text{Suma are 100 de termeni.}
\end{cases}\\\\\\
\text{Folosim formula: }\\\\
S_n=b_1\cdot \frac{q^n-1}{q-1} \\\\\\\\
S_{100}= \frac{1}{5} \cdot \frac{ \left(\dfrac{1}{5}\right)^{100}-1 }{\dfrac{1}{5}-1} =\\\\\\
=\frac{1}{5} \cdot \frac{1- \left(\dfrac{1}{5}\right)^{100} }{1-\dfrac{1}{5}} =
[/tex]
[tex]\displaystyle\\
=\frac{1}{5} \cdot \frac{1- \left(\dfrac{1}{5}\right)^{100} }{\dfrac{4}{5}} =
\frac{1}{5}\cdot \dfrac{5}{4}\cdot \left( 1- \left(\dfrac{1}{5}\right)^{100} \right)=\\\\\\
=\dfrac{1}{4}\cdot \left( 1- \left(\dfrac{1}{5}\right)^{100} \right)=
\dfrac{1}{4}\cdot \left( 1- \dfrac{1}{5^{100}} \right)=
\dfrac{1}{4}\cdot \dfrac{5^{100}-1}{5^{100}} =\boxed{\bf\dfrac{5^{100}-1}{4\cdot5^{100}}}
[/tex]
