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

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...

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    $\begingroup$ It's fully correct... just expand the term in the parenthesis and cancel out the two terms in the middle... $\endgroup$
    – b00n heT
    Commented Jan 31, 2014 at 22:43
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    $\begingroup$ Multiply through. You get on top $1-q^{n+1}+q^{n+1}-q^{n+2}$. $\endgroup$
    – André Nicolas
    Commented Jan 31, 2014 at 22:43
  • $\begingroup$ I can't believe I didn't see that. I'm no good at this sort of thing. $\endgroup$
    – furashu
    Commented Feb 1, 2014 at 2:05
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    $\begingroup$ also: what happens if q= 1 in this sum? obviously the zero denominator causes a problem $\endgroup$
    – furashu
    Commented Feb 1, 2014 at 2:13

2 Answers 2

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

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

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  • $\begingroup$ Can you please help me with this one? math.stackexchange.com/questions/2177377/… $\endgroup$
    – HKT
    Commented Mar 8, 2017 at 11:33
  • $\begingroup$ I don't know how to factor it ending up with $\frac{x^{k+1}-1}{x-1}$. $\endgroup$
    – HKT
    Commented Mar 8, 2017 at 11:34

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