How do i prove this problem with a product and its fraction part?

Question

How do I prove that $\forall \:\:r\:\in \mathbb{R}$,$\:r\:>\:1,\:\sum _{i=0}^n\:xr^i=x\frac{1-r^{n+1}}{1-r}$.

Answer 1

Typing off my phone here. Let’s prove it by induction. It is obviously true for $n=1$ (base case)

Assume the equation you have is true at $n-1$, namely

$\:r\:>\:1,\:\sum _{i=0}^{n-1}\:xr^i=x\frac{1-r^{n}}{1-r}$

Now let’s prove it for $n$

$\sum _{i=0}^{n}\:xr^i=\underbrace{x\frac{1-r^{n}}{1-r} }_{\sum _{i=0}^{n-1}\:xr^i}+xr^n=x\frac{1-r^{n} + r^n -r^{n+1}}{1-r}= x\frac{1-r^{n+1}}{1-r} $

Answer 2

$(1-r)\sum_{i=0}^n xr^i = \sum_{i=0}^nxr^i - \sum_{i=0}^nxr^{i+1} = \sum_{i=0}^nxr^i-\sum_{i=1}^{n+1}xr^{i}$

$=xr^0+\sum_{i=1}^nxr^i-\sum_{i=1}^nxr^i-xr^{n+1}=x(1-r^{n+1}).$

Divide both sides by $1-r$ and you're done.