Show that $\sum_{i=0}^{2n} (-1)^ix^i = \frac{x^{2n+1}+1}{x+1}$

Question

How can one show that if $x\neq-1$:

$$\sum_{i=0}^{2n} (-1)^ix^i = \frac{x^{2n+1}+1}{x+1}$$

I know that:

$$\sum_{i=1}^{n} r^{i-1} = \frac{r^n-1}{r-1}$$

But then how to proceed?

Answer 1

HINT

Note that $$\sum_{i=0}^{2n} (-1)^ix^i =\sum_{i=0}^{2n} (-x)^i = \frac{(-x)^{2n+1}-1}{-x-1} = \frac{(x)^{2n+1}+1}{x+1}$$