Question

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

Answer 1

Given:

`\begin{gathered} a_1=6 \\ r=3 \\ n=8 \end{gathered}`

To find: The sum of the geometric series

Since,

`r=3>1`

The sum formula for the geometric series is,

`S_n=(a_1(r^n-1))/(r-1)`

Substituting the given values, we get

`\begin{gathered} S_n=(6(3^8-1))/((3-1)) \\ =(6(6561-1))/(2) \\ =3(6560) \\ =19680 \end{gathered}`

Thus, the sum of the given series for 8 terms is 19680.

Final answer: The sum is,

`S_n=19680`