# How do you find sum of geometric series 5+20+80+...+ 20480?

Feb 27, 2016

$5 + 20 + 80 + \ldots + 20480$ can be written as, $5 \setminus \times 1 + 5 \setminus \times {4}^{1} + 5 \setminus \times {4}^{2} + \ldots + 5 \setminus \times {4}^{6}$
$= 5 \setminus \times \left(1 + 4 + {4}^{2} + \ldots + {4}^{6}\right) = 5 \setminus \times \left(\setminus \frac{{4}^{7} - 1}{4 - 1}\right) = 27305$.

#### Explanation:

A geometric series with $N$ terms is of the general form :
${G}_{n} = a + a {x}^{1} + a {x}^{2} + \ldots + a {x}^{n}$

Let ${G}_{n} = a \setminus \times \left(1 + {x}^{1} + {x}^{2} + \ldots + {x}^{n}\right) = a . {S}_{n}$ where
${S}_{n} = 1 + {x}^{1} + {x}^{2} + \ldots + {x}^{n}$;
There are two ways of writing ${S}_{n}$,

${S}_{n} = 1 + x \setminus \times \left(1 + {x}^{1} + {x}^{2} + \ldots + {x}^{n - 1}\right)$
$\setminus q \quad = 1 + x {S}_{n - 1}$ ...... (Eq.1)

${S}_{n} = \left(1 + {x}^{1} + {x}^{2} + \ldots + {x}^{n - 1}\right) + {x}^{n} = {S}_{n - 1} + {x}^{n}$ ...... (Eq.2)

Since the LHS of these two equations are the same we can equate their RHS.

$1 + x {S}_{n - 1} = {S}_{n - 1} + {x}^{n} \setminus q \quad \setminus \rightarrow \left(x - 1\right) {S}_{n - 1} = \left({x}^{n} - 1\right)$
${S}_{n - 1} = \left(\setminus \frac{{x}^{n} - 1}{x - 1}\right)$
Therefore, ${S}_{n} = \left(\setminus \frac{{x}^{n + 1} - 1}{x - 1}\right)$

This Problem: a=5; \qquad x=4; \qquad n=6
${G}_{6} = a . {S}_{6} = 5 \setminus \times \left(\setminus \frac{{4}^{7} - 1}{4 - 1}\right) = 27305$.