Proof by Induction: $\sum_0^nx^i=(1-x^{n+1})/(1-x)$ I'm trying to do my Maths assignment but I can't get this done.
I can do other questions but this one is different.
This is a picture of the question.

(source: gyazo.com) 
My usual first step is to proof it when n = 1. But the problem is, that I don't know what x is. It says x is not equal to 1, but I still don't know what to do next.
Thank you.
 A: It doesn't matter what $x$ is, as long as it isn't $1$ (so $1-x$ isn't $0$, and you can divide by it). Just treat it as a normal inductive proof:
Suppose $\sum_{i=0}^{n}{x^{i}}=\frac{1-x^{n+1}}{1-x}$. Then
$$\sum_{i=0}^{n+1}{x^{i}}=\sum_{i=0}^{n}{x^{i}}+x^{n+1}=\frac{1-x^{n+1}}{1-x}+x^{n+1}=\frac{(1-x^{n+1})+(1-x)x^{n+1}}{1-x}$$
$$=\frac{1-x^{n+2}}{1-x},$$
which completes the inductive step.
The base case $n=0$ is LHS=$x^{0}+x^{1}=1+x$ (no matter what $x$ is), RHS=$\frac{1-x^{2}}{1-x}=1+x$, so it's true for all $n$.
A: You need to let $x$ be an arbitrary number, which is just not $1$. 
For the base case ($n = 0$): 
We have $$x^0 = 1 = \frac{1-x}{1-x} = \frac{1-x^{0+1}}{1-x}$$
So the theorem holds in the base case. 
For the inductive step: 
$$\sum_{i=0}^{n+1} x^i = \sum_{i=0}^{n} x^i + x^{n+1}$$
And by induction we know $\sum_{i=0}^n x^i = \frac{1 - x^{n+1}}{1-x}$, so plugging this in we get $$\sum_{i=0}^{n+1}x^i = \frac{1-x^{n+1}}{1-x} + x^{n+1} = \frac{1-x^{n+1}}{1-x} + \frac{(1-x)x^{n+1}}{1-x} = \frac{1 - x^{n+2}}{1-x}.$$
So by induction we're done. 
One can also note that there is another proof, without induction that is neater. 
$$(\sum_{i=0}^{n} x^i) (1-x) = \sum_{i=0}^{n} x^i - \sum_{i=0}^{n} x^{i+1} = 1 - x^{n+1}$$ because all the terms in the middle cancel out.
A: Induction involving $x$ is no different from other types of induction. Just treat $x$ normally and we can still prove it for the general case.
Let $$P_n:=\sum_{i=0}^nx^i=\frac{1-x^{n+1}}{1-x}$$
Then for the base case i.e. $n=0$ we have
$$\sum_{i=0}^0x^i=\frac{1-x^{1+0}}{1-x}$$
$$1=\frac{1-x}{1-x}$$
$$1=1$$
He have esablished the RHS=LHS and so $P_0$ is true.
Now assume that the statement is true for n. 
$$\sum_{i=0}^nx^i=\frac{1-x^{n+1}}{1-x}$$
We must show that it is also true for $n+1$. 
For the RHS we have
$$RHS=\frac{1-x^{(n+1)+1}}{1-x}=\frac{1-x^{n+2}}{1-x}$$
And for the LHS we have
$$LHS=\sum_{i=0}^{n+1}x^i=\sum_{i=0}^{n}x^i+x^{n+1}=\frac{1-x^{n+1}}{1-x}+x^{n+1}=\frac{1-x^{n+1}}{1-x}+\frac{x^{n+1}(1-x)}{(1-x)}$$
$$=\frac{1-x^{n+1}+x^{n+1}(1-x)}{1-x}=\frac{1-x^{n+1}+x^{n+1}-x^{n+2}}{1-x}=\frac{1-x^{n+2}}{1-x}$$
Since LHS=RHS $P_{n+1}$ is true. The rest follows.
A: For $n=1$, the right hand side is
$$\begin{align*}
\frac{1 - x^2}{1-x} 
 &= \frac{(1-x)(1+x)}{1-x} \\
 &= 1+x, \quad x \ne 1
\end{align*}$$
which is identical to the right hand side.
