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I'm having some issues trying to understand a problem. I think that what makes me confused is that we have to put the k+1 in the exponent of a fraction, at the beginning of the induction step, where k=k+1. From that point on I'm not really sure how to continue with this proof.

Here's the problem at issue:

Use mathematical induction on $n$ to prove that $\forall n \in \mathbb{N}$: $$ \sum_{i=1}^n \left( \frac{1}{3} \right)^i = \frac{1}{2} \cdot \frac{3^n -1}{3^n} $$

Can anybody give me a hand?

Thanks in advance!

Matti P.
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youser202342
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  • So can you show us the expression that you wrote, and that's making you confused? I edited the question using MathJax and you can continue the same way. – Matti P. Nov 07 '19 at 11:49
  • First you can check the case $n=1$. In that case you get $\left(\frac{1}{3}\right)^1=\frac{1}{2}\cdot\frac{3^1-1}{3^1}$, which you can see is equivalent to $\frac{1}{3}=\frac{1}{3}$. Next you assume that the formula is true for an arbitrary value $n$ and try to prove the same formula for the next value $n+1$. Note that the left hand side for $n+1$ is $\sum_{i=1}^{n+1}\left(\frac{1}{3}\right)^i=\sum_{i=1}^{n}\left(\frac{1}{3}\right)^i + \left(\frac{1}{3}\right)^{n+1}=\frac{1}{2}\cdot\frac{3^n-1}{3^n}+\frac{1}{3^{n+1}}$. So, it only remains to prove that – conditionalMethod Nov 07 '19 at 11:56
  • $\frac{1}{2}\cdot\frac{3^n-1}{3^n}+\frac{1}{3^{n+1}}=\frac{1}{2}\cdot\frac{3^{n+1}-1}{3^{n+1}}$. – conditionalMethod Nov 07 '19 at 11:57
  • And that's exactly where I get confused. I tried different approaches, like getting the common divisor, but it didn't lead to anything. How should I continue? I feel like I'm that close to grasp this proof, but a little detail is holding me back :( – youser202342 Nov 07 '19 at 12:04
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    Hint: $$ \begin{split} \frac{1}{2} \cdot \frac{3^n -1}{3^n} + \frac{1}{3^{n+1}} &= \frac{1}{2} \cdot \frac{3}{3} \cdot \frac{3^n -1}{3^n} + \frac{2}{2} \cdot \frac{1}{3^{n+1}} \ &= \frac{1}{2} \cdot \frac{3^{n+1}-3}{3^{n+1}} + \frac{1}{2}\cdot \frac{2}{3^{n+1}} \end{split} $$ – Matti P. Nov 07 '19 at 12:08
  • Ahh, it clicked, thanks a lot! I see why you multiply by 3 and 2, then all we have to do is take one half as common factor and multiply the other two fractions to get the right result. – youser202342 Nov 07 '19 at 12:45

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