Question

How do you find the sum of the finite geometric sequence of `Sigma 5(-3/2)^(n-1)` from n=1 to 8?

Answer 1

`-6305/128`

Explanation:

In general for finite geometric series:

`sum_(k=1)^(n)(a*r^(k-1)) = (a*(1-r^(n)))/(1-r)`

In this case `a=5`, `r=-3/2`, and `n=8`, so:

`sum_(k=1)^(n)(a*r^(k-1)) = (5*(1-(-3/2)^(8)))/(1-(-3/2))=-6305/128`