How do you sum the series 1+a+a^2+a^3+...+a^n?

Jul 3, 2015

$1 + a + {a}^{2} + {a}^{3} + \ldots + {a}^{n} = \frac{1 - {a}^{n + 1}}{1 - a}$

Explanation:

$\left(1 + a + {a}^{2} + {a}^{3} + \ldots + {a}^{n}\right) \left(1 - a\right) = 1 - {a}^{n + 1}$

Divide both sides by $\left(1 - a\right)$ to get:

$1 + a + {a}^{2} + {a}^{3} + \ldots + {a}^{n} = \frac{1 - {a}^{n + 1}}{1 - a}$

Jul 3, 2015

The solution George C.gives is very elegant. Here's another that has its own uses.

Explanation:

Let

${S}_{n} = 1 + a + {a}^{2} + {a}^{3} + \cdot \cdot \cdot + {a}^{n}$

Multiply by $a$ to get:

$a {S}_{n} = \text{ } a + {a}^{2} + {a}^{3} + \cdot \cdot \cdot + {a}^{n} + {a}^{n + 1}$

Now subtract:

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

So, again:

${S}_{n} = \frac{1 - {a}^{n + 1}}{1 - a}$