TITLE: Proof of the formula $1+x+x^2+x^3+ \cdots +x^n =\frac{x^{n+1}-1}{x-1}$
QUESTION [5 upvotes]: Possible Duplicate:
Value of $\sum x^n$ 

Proof to the formula 
$$1+x+x^2+x^3+\cdots+x^n = \frac{x^{n+1}-1}{x-1}.$$

REPLY [9 votes]: Since $1-x^{n+1}$ has $1$ as a root, the quotient $\frac{1-x^{n+1}}{1-x}$ is a polynomial.
If $\mathbb F_q$ is a finite field with $q$ elements and $V$ is a $\mathbb F_q$-vector space of dimension $n+1$, then $\frac{1-q^{n+1}}{1-q}=|P(V)|$ is the cardinal of the projective space attached to $V$. Now $P(V)$ can be described as a disjoint union $$P(V)=\mathbb A^0\sqcup\mathbb A^1\sqcup \mathbb A^2\sqcup\cdots\sqcup\mathbb A^n$$ where $\mathbb A^k$ is, for each $k$, an affine space of dimension $k$ over $\mathbb F_q$ (which is a complicated way of saying, as far as our purposes go, a vector space over $\mathbb F_q$ of dimension $k$) Since $|\mathbb A^k|=q^k$, we find that
$$\frac{1-q^{n+1}}{1-q}=1+q+q^2+q^3+\cdots+q^n$$
for all numbers $q$ which are powers of prime numbers. It follows that
$$\frac{1-x^{n+1}}{1-x}=1+x+x^2+x^3+\cdots+x^n$$
as polynomials, because the equality holds for infinitely many values of $x$ (and we are working over $\mathbb Z$...)