Proving the geometric sum formula by induction

Question

$$\sum_{k=0}^nq^k = \frac{1-q^{n+1}}{1-q}$$

I want to prove this by induction. Here's what I have.

$$\frac{1-q^{n+1}}{1-q} + q^{n+1} = \frac{1-q^{n+1}+q^{n+1}(1-q)}{1-q}$$

I wanted to factor a $q^{n+1}$ out of the second expression but that 1- is screwing it up...

Answer 1

$$1 - q^{n+1} + q^{n+1}(1-q) = 1 - q^{n+1}(1 - (1-q)) = 1 - (q^{n+1} \cdot q) = \cdots $$

Answer 2

Did you try expanding the numerator? You have $1-q^{n+1}+q^{n+1}-q^{n+2}$..