Sum of the k-th powers of x for k = 0 to n

by RoRi

July 5, 2015

Prove that
$\sum_{k=0}^n x^k = \frac{1-x^{n+1}}{1-x} \qquad \text{for } x \neq 1.$

Proof. We use properties of finite sums to compute,

$(1-x) \sum_{k=0}^n x^k = \sum_{k=0}^n (x^k - x^{k+1}) = - \sum_{k=0}^n (x^{k+1} -x^k) = -(x^{n+1} - 1) = 1-x^{n+1}.$

The second to last inequality follows from the telescoping property. But then solving for the sum we are interested in, we have

$\sum_{k=0}^n x^k = \frac{1-x^{n+1}}{1-x}. \qquad \blacksquare$
If $x = 1$ we have (using this, and the fact that $1^k = 1$ for $k = 0, \ldots, n$ )
$\sum_{k=0}^n x^k = \sum_{k=0}^n 1 = n+1.$