Inductive step in the induction: $\sum^{n}_{i=0} q^i = \frac {1-q^{n+1}}{1-q}\times2$ I am trying induction for the following formula:
$$\sum^{n}_{i=0} q^i = \frac {1-q^{n+1}}{1-q}\times2$$
I have done the initial step which gives me for $n=1$ for both sites $1+q$
In the inductive step I wrote:
$$(\sum^{n}_{i=0} q^i ) \times \frac {1-q^{n+1}}{1-q}\times2 =\frac {1-q^{n+1}}{1-q}\times2 \times \frac {1-q^{n+2}}{1-q}\times2 = 4 \times \frac {-q*q^n - q^2 q^n + 1 + q^3 q^{2n}}{1-q}$$
However, here I am stuck...
I really would appreciate your answer!!!
 A: Beginning note:  all operations below assume that $q\ne 1$.  To avoid other pitfalls, also assume that $q\ne 0$.
For induction on a sum, you would use the following process:
$$\sum^{n}_{i=0} q^i = \frac {1-q^{n+1}}{1-q}\times2$$


*

*Show for $n=1$ (or $n=0$, if desired)

*Assume statement is true for an arbitrary $n=k\gt 1$

*Show statement holds for $n=k+1$


For the base case, use $n=1$ as the point where you started.  Then we have:
$$\sum^{1}_{i=0} q^i = \frac {1-q^{1+1}}{1-q}\times2 \implies 1+q=\frac {(1-q)(1+q)}{1-q}$$
$$=\frac {1-q^2}{1-q}\ne \frac {1-q^2}{1-q}\times 2$$
Already, we have failed to show the required induction.  At this point I would start again, ensuring that your formula is correct.
Continuing forward as if we had succeeded at the base step, we would have the following:
Show that:
$$\sum^{k+1}_{i=0} q^i = \frac {1-q^{k+2}}{1-q}\times2$$
To do this, we would use
$$\sum^{k}_{i=0} q^i = \frac {1-q^{k+1}}{1-q}\times2$$
(our assumption step) and add the $k+1$ term to both sides, like so:
$$\sum^{k}_{i=0} q^i + q^{k+1} = \frac {1-q^{k+1}}{1-q}\times2 + q^{k+1}$$
$$\implies \sum^{k+1}_{i=0} q^i = \frac {1-q^{k+1}}{1-q}\times2+\frac{(1-q)q^{k+1}}{1-q}$$
$$=\frac {2-q^{k+1}-q^{k+2}}{1-q}\ne \frac {1-q^{k+2}}{1-q}\times2$$
The problems we found at the base case continue through the entire induction process and show again that the original formula as stated cannot be correct.
A related form of your formula looks like this in the final steps:
$$\sum^{k+1}_{i=0} q^i = \frac {1-q^{k+1}}{1-q}+\frac{(1-q)q^{k+1}}{1-q}$$
$$=\frac {1-q^{k+1}+(1-q)q^{k+1}}{1-q}=\frac {1-q^{k+2}}{1-q}$$
