# Proving the geometric sum formula by induction

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

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...

• It's fully correct... just expand the term in the parenthesis and cancel out the two terms in the middle... – b00n heT Jan 31 '14 at 22:43
• Multiply through. You get on top $1-q^{n+1}+q^{n+1}-q^{n+2}$. – André Nicolas Jan 31 '14 at 22:43
• I can't believe I didn't see that. I'm no good at this sort of thing. – furashu Feb 1 '14 at 2:05
• also: what happens if q= 1 in this sum? obviously the zero denominator causes a problem – furashu Feb 1 '14 at 2:13

$$1 - q^{n+1} + q^{n+1}(1-q) = 1 - q^{n+1}(1 - (1-q)) = 1 - (q^{n+1} \cdot q) = \cdots$$
Did you try expanding the numerator? You have $1-q^{n+1}+q^{n+1}-q^{n+2}$..
• I don't know how to factor it ending up with $\frac{x^{k+1}-1}{x-1}$. – HKT Mar 8 '17 at 11:34