Through simple induction, I would like to prove the following:
$\forall x \in \mathbb{N^{>1}}, \forall y \in \mathbb{N^{>0}}, \sum\limits_{i=0}^{y}x^i = \frac{x^{y+1} - 1}{x -1}$
Let $P(y, x)$ represent $ \sum\limits_{i=0}^{y}x^i = \frac{x^{y+1} - 1}{x -1}$
Base Case: $P(y=1, x=2)$
$\sum\limits_{i=0}^{1}2^i = \frac{2^{1+1} - 1}{2 -1}$
$3=3$
Therefore the base case holds
Inductive Step: ?
I believe this type of problem is called multidimensional induction, and I tried reading some websites about this topic, but I can't comprehend the inductive step. If someone could guide me in the right direction, that would be good. Thanks
Induction is not necessary for this proof, and it holds for all $x\in\mathbb C\setminus\{1\}$. Let $S(y) = \sum_{i=0}^y x^i$. Then $$ (1-x)S(y) = (1-x)\sum_{i=0}^y x^i = \sum_{i=0}^y x^i - \sum_{i=0}^y x^{i+1} = 1 - x^{y+1}. $$ Dividing by $(1-x)$, we conclude that $$ S(y) = \frac{1-x^{y+1}}{1-x}. $$
It is possible to prove this by induction on $y$, but that is actually a more complicated proof than this trick. Our base case is $y = 0$. Then $\sum_{i=0}^0 x^i = 1 = \frac{1-x^{0+1}}{1-x}$. Now assume that for some nonnegative integer $y$ that $\sum_{i=0}^y x^i = \frac{1-x^{y+1}}{1-x}$. Then \begin{align} \sum_{i=0}^{y+1} x^i &= \sum_{i=0}^y x^i + x^{y+1}\\ &= \frac{1-x^{y+1}}{1-x} + x^{y+1}\\ &= \frac{1-x^{y+1} +(1-x)x^{y+1}}{1-x}\\ &= \frac{1 - x^{y+1} + x^{y+1} - x^{y+2}}{1-x}\\ &= \frac{1 - x^{y+2}}{1-x}, \end{align} so that the result holds for all nonnegative integers $y$.