Please help me find the sum of this series: a1: 6, r=3, n=8

Given:
[tex]\begin{gathered} a_1=6 \\ r=3 \\ n=8 \end{gathered}[/tex]To find: The sum of the geometric series
Since,
[tex]r=3>1[/tex]The sum formula for the geometric series is,
[tex]S_n=\frac{a_1(r^n-1)}{r-1}[/tex]Substituting the given values, we get
[tex]\begin{gathered} S_n=\frac{6(3^8-1)}{(3-1)} \\ =\frac{6(6561-1)}{2} \\ =3(6560) \\ =19680 \end{gathered}[/tex]Thus, the sum of the given series for 8 terms is 19680.
Final answer: The sum is,
[tex]S_n=19680[/tex]