Question

Given the series 1+2+4+8+.... find the 8th term​

Answer 1

Consider the
`n`-th partial sum of a geometric series with first term
`a` and common ratio
`r` between consecutive terms. Then the sum of the first
`n` terms of this series is

`S = a + ar + ar^2 + \cdots + ar^(n-1)`

Multiply both sides by
`r`, then subtract
`rS` from
`S` and solve for
`S`.

`rS = ar + ar^2 + ar^3 + \cdots + ar^n`

`S - rS = a - ar^n`

`\implies S = (a(1-r^n))/(1-r)`

In the given series, we have
`a=1` and
`r=2`; then the sum of the series with
`n=8` terms is

`S = (1-2^8)/(1-2) = 2^8 - 1 = \boxed{255}`