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Prove by induction that for any real number $q≠1$ and any $n\in \mathbb N$ we have $ \sum_{i=0}^n q^i=\frac{q^{n+1}-1}{q-1} $

Martin Sleziak
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LifeFlow
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  • this is abuse. You don't need induction here. $$\sum_{i=0}^n q^i=\frac{q^{n+1}-1}{q-1}$$ is the simplest expression ever. – Guy Mar 06 '14 at 06:47
  • @JoseAntonio nothing apparently. – Guy Mar 06 '14 at 06:47
  • What is your question? Just show the formula for $n=0$ and compute the case $n+1$ from $n$, $\dfrac{q^{n+1}-1}{q-1}+q^{n+1}=\dfrac{q^{n+1}-1}{q-1}+\dfrac{q^{n+1}(q-1)}{q-1}$ – Blah Mar 06 '14 at 06:47
  • I'm going to be honest I don't really know where to start. I know that you have to prove the base case, which I would assume would be 0. – LifeFlow Mar 06 '14 at 06:49
  • @Sabyasachi : I don't think is an abuse, it is important to realise that manipulation of any kind in finite series are ultimately justified through the axiom of induction. – Jose Antonio Mar 06 '14 at 06:50
  • @JoseAntonio ultimately justified through induction?

    $$S = 1+ q + q^2 +q^3 + \cdots + q^n$$

    $$qS= q+ q^2 +q^3 + \cdots + q^n + q^{n+1}$$

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

    I might be wrong, so this is a humble question, where is the axiom of induction used here?

     – Guy Mar 06 '14 at 06:53
  • @Sabyasachi: For a foundational course, it is important to make sure that you can in fact supply a rigorous proof for any statement which is intuitively obvious, in this case of finite series, induction justify all, and also it is important to see that this is not directly available for the infinite series. – Jose Antonio Mar 06 '14 at 07:00
  • @JoseAntonio Okay I see. Although to be honest, I never liked the whole emphasis on rigor. It is obvious if it is obvious. Although without calculus, the infinite series isn't readily available. – Guy Mar 06 '14 at 07:03
  • @Sabyasachi: For example the Fubini's theorem for finite series it's obvious, but its proof is not trivial, at least for me, only by induction I could give it a justification. – Jose Antonio Mar 06 '14 at 07:09
  • @JoseAntonio I was referring to cases in which the actual proof is obvious. For instance, in this case the geometric series. I understand what you mean by induction is important then, if you suspect a theorem is valid, but can't prove it, induction is almost always the easy way. – Guy Mar 06 '14 at 07:16

1 Answers1

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Base Case:

For $n=0$, our sum is $\sum_{i=0}^n q^i$ is equal to 1, which is also the value of $\frac{q^{n+1}-1}{q-1}$ at $n=0$. Hence base case justified.

Induction hypothesis:

Let

$$\sum_{i=0}^k q^i=\frac{q^{k+1}-1}{q-1}$$ for some $k \ge0$.

Inductive step:

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

Taking it to a common denominator, and some simple algebraic manipulation.

$$\sum_{i=0}^{k+1} q^i=\frac{q^{k+2}-1}{q-1}$$

Q.E.D

Guy
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