# What is the formula for the sum of a geometric sequence?

Sep 2, 2015

The formula is: ${S}_{n} = \frac{{a}_{1} \left(1 - {q}^{n}\right)}{1 - q}$

#### Explanation:

To proove this formula you have first to write the sum of a geometric sequence:

${S}_{n} = {a}_{1} + {a}_{1} \cdot q + {a}_{1} \cdot {q}^{2} + \ldots {a}_{1} \cdot {q}^{n - 1}$

${S}_{n} = {a}_{1} \cdot \left(1 + q + {q}^{2} + \ldots {q}^{n}\right)$

We can multiply the last equality by $\left(1 - q\right)$
We get:

$\left(1 - q\right) \cdot {S}_{n} = {a}_{1} \cdot \left(1 - q\right) \cdot \left(1 + q + {q}^{2} + \ldots {q}^{n}\right)$

The right hand side can be written as ${a}_{1} \cdot \left({1}^{n} - {q}^{n}\right)$ or
${a}_{1} \cdot \left(1 - {q}^{n}\right)$. So finally we get:

$\left(1 - q\right) \cdot {S}_{n} = {a}_{1} \cdot \left(1 - {q}^{n}\right)$

Supposing $q \ne 1$, we can divide both sides by $1 - q$ and get:

${S}_{n} = \frac{{a}_{1} \left(1 - {q}^{n}\right)}{1 - q}$

That concludes the proof.