Question

How do you find the sum of `4(3/2)^(i-1)` from i=0 to 6?

Answer 1

`2059/24`

Explanation:

This is a geometric series `sum_{i=0}^n ar^i` with first term

`a = 4(3/2)^{-1}=8/3`

and common ratio `r = 3/2`

The sum is

`um_{i=0}^n ar^i = a{r^{n+1}-1}/{r-1} = 8/3 {(3/2)^7-1}/{3/2-1} = 8/3 {2187/128-1}/{1/2} = 16/{3\times 128}(2187-128)= 2059/24`