Help with a step in rearranging this problem i'm working through the proof of this theorem;
If $x$ is any real number other than $1$, then $$\sum_{j = 0}^{n -1} x^j = 1 + x + x^2 + \cdots + x^{n-1} = \frac{x^n-1}{x-1}$$
But i'm struggling with an intermediate step, so I hope you can help. Please don't take any large jumps in rearranging and also I would appreciate it if you would leave the last steps to the completed theorem for me to continue working on.
So here's where i'm stuck
$$\frac{(x^k) - 1}{ x - 1} + x^k = \frac{x^k - 1 + x^{k+1} - x^k}{x - 1}$$
I just can't seem to get LHS to equal the RHS
 A: Put each term over a common denominator, just like you would if you were adding two rational numbers. Since $\frac{x-1}{x-1}=1$ for any $x\neq 1$, we have
$$x^k \;=\;x^k\cdot 1\;=\;x^k\cdot\frac{x-1}{x-1}\;=\; \frac{x^k\cdot (x-1)}{x-1}\;=\;\frac{x^{k+1}-x^k}{x-1}$$
and therefore
$$\frac{x^k-1}{x-1}+x^k=\frac{x^k-1}{x-1}+\frac{x^{k+1}-x^k}{x-1}=\frac{(x^k-1)+(x^{k+1}-x^k)}{x-1}.$$
A: Here's an alternate proof for the formula of a finite geometric series. Define:
$$
S = \sum_{j = 0}^{n -1} x^j = 1 + x + x^2 + \cdots + x^{n-1}
$$
Then multiplying both sides by $x$ yields:
$$
xS = x+x^2+\cdots+x^n
$$
Subtracting the first equation from the second yields:
$$ \begin{array}{}
&[xS&=&  & &x&+&x^2&+&\cdots&+&x^{n-1}&+&x^n&]\\
-&[S &=&1&+&x&+&x^2&+&\cdots&+&x^{n-1}&&&]\\
\hline
&[xS-S &=& -1&&&&&&&&&+&x^n&]\\
\end{array}$$
Hence, we have:
$$ \begin{align*}
xS-S&=-1+x^n\\
S(x-1)&=x^n-1\\
S &= \dfrac{x^n-1}{x-1}
\end{align*}$$
as desired.
A: A slight modification of Adriano's answer: add $x^{n}$ to both sides, then factor out $x$:
$$
S_{n}+x^{n}=1+xS_n
$$
hence you get what you need after rearranging terms
