Question

How do you find the sum of the geometric series `Sigma 64(3/4)^(n-1)` n=1 to 8?

Answer 1

`S_8=(4^8-3^8)/4^7`

Explanation:

The sum of the first `n` terms of a geometric series with general term `b_m` and common ratio `q` is

`S_n = b_1 * (q^n-1)/(q-1)`

For `q<1`, this is usually written as

`S_n = b_1 (1-q^n)/(1-q)`

`b_n = 64(3/4)^(n-1)=> {(b_1 = 64),(q=b_(n+1)/b_n = 3/4) :}`

`S_8 = 64*(1-(3/4)^8)/(1-3/4)=(4^8-3^8)/4^7`